If you are familiar with Lie algebras and representation theory, then you can skip my introduction and get right to the two questions (conjectures) at the end. You might only need to read the next paragraph where I define $h$, $x$, and $y$.
Recall that $\mathfrak{sl}_2(\mathbb{C})$ is the Lie algebra of $2$-by-$2$ traceless matrices over $\mathbb{C}$. Let $\{h,x,y\}$ be an $\mathfrak{sl}_2(\mathbb{C})$-triple, that is, $\{h,x,y\}$ is a basis of the three dimensional $\mathbb{C}$-vector space $\mathfrak{sl}_2(\mathbb{C})$ such that $$[x,y]=h\,,\,\, [h,x]=+2x\,,\text{ and }[h,y]=-2y\,,$$ where $[\_,\_]$ is the Lie bracket of $\mathfrak{sl}_2(\mathbb{C})$. With a proper change of basis, it suffices to say that $$h=\begin{bmatrix}+1&0\\0&-1\end{bmatrix}\,,\,\,x=\begin{bmatrix}0&1\\0&0\end{bmatrix}\,,\text{ and }y=\begin{bmatrix}0&0\\1&0\end{bmatrix}\,.$$
A (left) representation $V$ of $\mathfrak{sl}_2(\mathbb{C})$ is given. That is, $V$ is a $\mathbb{C}$-vector space equipped with a unitary left $\mathfrak{U}\big(\mathfrak{sl}_2(\mathbb{C})\big)$-module structure, where $\mathfrak{U}\big(\mathfrak{sl}_2(\mathbb{C})\big)$ is the universal enveloping algebra of $\mathfrak{sl}_2(\mathbb{C})$. Hence, for $a,b\in \mathfrak{sl}_2(\mathbb{C})$ and $v\in V$, $$[a,b]\cdot v=a\cdot(b\cdot v)-b\cdot(a\cdot v)\,,$$ where $\cdot:\mathfrak{sl}_2(\mathbb{C})\times V\to V$ is the $\mathfrak{sl}_2(\mathbb{C})$-action on $V$.
An example of a representation of $\mathfrak{sl}_2(\mathbb{C})$ is given by $V:=\mathbb{C}[s,t]$, in which $x$, $y$, and $h$ act on $\mathbb{C}[s,t]$ as follows: $$(x\cdot f)(s,t):=s\,\frac{\partial}{\partial t}\,f(s,t)$$ $$(y\cdot f)(s,t):=t\,\frac{\partial}{\partial s}\,f(s,t)$$ and $$(h\cdot f)(s,t):=s\,\frac{\partial}{\partial s}\,f(s,t)-t\,\frac{\partial}{\partial t}\,f(s,t)\,,$$ for all $f(s,t)\in \mathbb{C}[s,t]$. Note that the subspace $\mathbb{C}_k[s,t]$ of $\mathbb{C}[s,t]$ of homogeneous polynomials of degree $k\in\mathbb{Z}_{\geq 0}$ is a $(k+1)$-dimensionl irreducible representation of $\mathbb{C}[s,t]$, and $$\mathbb{C}[s,t]=\bigoplus_{k=0}^\infty\,\mathbb{C}_k[s,t]$$ is a decomposition of $\mathbb{C}[s,t]$ into a direct sum of irreducible representations. Every finite-dimensional irreducible representation of $\mathfrak{sl}_2(\mathbb{C})$ is isomorphic to some $\mathbb{C}_k[s,t]$. Observe that $x^{k+1}$ and $y^{k+1}$ are the zero operator on $\mathbb{C}_k[s,t]$, but $x^k$ and $y^k$ are nonzero. Thus, on $\mathbb{C}_k[s,t]$, $x$ and $y$ have the same nilpotency index, which is $k+1$.
Note also that the actions of $x$ and $y$ are both nilpotent on any finite-dimensional representation of $\mathfrak{sl}_2(\mathbb{C})$ (because such a representation can be written as a direct sum of finitely many finite-dimensional irreducible representations). The nilpotency index of $x$ equals the nilpotency index of $y$ in any finite-dimensional representation of $\mathfrak{sl}_2(\mathbb{C})$. We know that $x$ and $y$ do not need to act nilpotently (or locally nilpotently) on an infinite-dimensional representation (see here). However, what happens if $x$ or $y$ is a nilpotent operator? Does it mean that both are nilpotent operators, and when so, do their nilpotency indices conincide? Furthermore, what are all the representations of $\mathfrak{sl}_2(\mathbb{C})$ such that $x$ and $y$ are nilpotent operators? That is, is the following conjecture true?
Conjecture I. The action of $x$ on a representation $V$ of $\mathfrak{sl}_2(\mathbb{C})$ is nilpotent if and only if the action of $y$ on $V$ is also nilpotent. In such cases, the nilpotency index of $x$ is the same as the nilpotency index of $y$. Finally, if $p\in\mathbb{Z}_{>0}$ is the nilpotency index of $x$, $V$ is isomorphic to a direct sum of $\mathbb{C}_k[s,t]$, where $k\in\{0,1,2,\ldots,p-1\}$, with at least one copy of $\mathbb{C}_{p-1}[s,t]$ (the multiplicity of each $\mathbb{C}_k[s,t]$ may be any cardinal).
Recall that a locally nilpotent linear operator $\phi$ on $V$ is a linear operator such that, for each $v\in V$, there exists a positive integer $m(v)$ depending on $v$ such that $\phi^{m(v)}(v)=0$, where $\phi^{j}$ denotes the $j$-th iteration of $\phi$. We know also from here that $x$ and $y$ may or may not be locally nilpotent. More importantly, one of $x$ and $y$ may be locally nilpotent, and the other is not locally nilpotent at all. Nonetheless, what happens if both $x$ and $y$ are locally nilpotent? Is the next conjecture true?
Conjecture II. If both $x$ and $y$ are locally nilpotent operators on a representation $V$ of $\mathfrak{sl}_2(\mathbb{C})$, then $V$ is isomorphic to a direct sum of finite-dimensional irreducible representations $\mathbb{C}_k[s,t]$ of $\mathfrak{sl}_2(\mathbb{C})$, i.e., $V$ is an integrable $\mathfrak{sl}_2(\mathbb{C})$-module.
The $\mathfrak{sl}_2(\mathbb{C})$-module $V$ is integrable if it is an $h$-weight $\mathfrak{sl}_2(\mathbb{C})$-module such that $x$ and $y$ are locally nilpotent. We know that an integrable $\mathfrak{sl}_2(\mathbb{C})$ is a direct sum of finite-dimensional irreducible representations. However, in the conjecture above, I do not assume that $V$ is a weight module with respect to $h$.
Let me in turn write a conjecture. Let $P_n(X)$ be the polynomial $\prod_{k=-n}^n(X-k)$.
This is obvious for $n=1$, and for $n=2$ it is given by the following, obtained by plain computation: $$x^2y^2-y^2x^2+(y^2x^2+x^2y^2-2yx^2y)h=2(h^3-h).$$
This conjecture implies both of your conjectures (see below). Small cases should be obtainable with computer help, or with some patience. For instance, for $n=3$, I'd compute $y^3x^3$, $y^2x^3y$, $yx^3y^2$ to put them as combinations of monomials of the form $x^ay^ah^b$, so as to eliminate terms and try to get the desired $P_2(h)=h(h^2-1)(h^2-4)$.
In addition, computation for enough small values might lead to a reasonable expectation for a general formula.
First, two conjecture-free facts. Say that an $\mathfrak{sl}_2$-module is locally P if every finitely generated submodule is locally P. Say that it is binilpotent if both $x$ and $y$ are nilpotent.
(Clearly locally finite-dimensional implies locally binilpotent, and the conjecture implies the converse, see below.)
The proof consists in showing, by induction on $k$, that if $x^kv=0$, then the submodule generated by $v$ is binilpotent. Assume that the case $k=1$ is done, and assume by induction that it's proved up to $k-1$. If $x^kv=0$, then $x(x^{k-1}v)=0$, so by induction $x^{k-1}v$ is contained in a binilpotent submodule $W$. In $V/W$, $xv=0$ and hence $x$ is contained in a binilpotent submodule $W'/W$ of $V/W$. Since both $W'/W$ and $W$ are binilpotent, $W'$ is nilpotent.
Now the case $k=1$. Let $W_k$ be the subspace of $V$ generated by the elements $y^kh^\ell v$ when $\ell$ ranges over $\mathbf{N}=\{0,1,2,\dots\}$; define $V_{\le k}=\sum_{i=0}^kW_{i}$.
One has (in the enveloping algebra) the equalities $$xy^n=y^nx+ny^{n-1}(h-n+1)\;(n\ge 1);\quad xh^n=(h-2)^nx\;(n\ge 0).$$
Hence, for $k\ge 1$, $$xy^kh^\ell v=(y^kx+ky^{k-1}(h-k+1))v=ky^{k-1}v-k(k-1)y^{k-1}hv;\quad xh^\ell v=(h-2)^\ell xv=0$$
Hence $xW_0=0$, $xW_k\subset W_{k-1}$, and clearly $yW_k\subset W_{k+1}$. Hence $W=\bigcup_{k\ge 0}W_k$ is a submodule.
Moreover, using that $y^kh^\ell=(h+2k)^\ell y^k$, we see, if $y^mv=0$, that $W=\bigcup_{0\le k<m}W_k$, and $x^m=y^m=0$ on $W$, so $W$ is binilpotent.
The next fact, probably very standard, is the following
Here is the short proof: let $V_n$ be the sum of all $(n+1)$-dimensional irreducible submodules. Then $V=\bigoplus_{n\ge 0}V_n$. So we can suppose that $V=V_n$. Let $I_n$ (in the enveloping algebra $U$) be the kernel of the irreducible $(n+1)$-dimensional representation. Then $U/I_n$ is isomorphic to the $(n+1)$-dimensional matrix algebra. Since the latter is a semisimple artinian ring, all its modules are projective, and hence every module is a direct sum of cyclic submodules, and in this case the only cyclic module is the $(n+1)$-dimensional one.
(Actually the proof adapts to arbitrary semisimple Lie algebras.)
Now assume my conjecture above. We work in the algebra $A$ generated by the operators in the representation, which is thus a quotient of the enveloping algebra. If $x$ is nilpotent in $A$, then $P_n(h)=0$ for some $n$, so we have the eigenvalue decomposition $V=\bigoplus_{k=-n}^nV_k$, where $V_k$ is the kernel of $h-k$, and $xV_k\subset V_{k+2}$, $yV_k\subset V_{k-2}$ for all $k$, and hence $x^{n+1}=y^{n+1}=0$ in $A$.
Next, in this setting, we can pick an eigenvector $v$ of $h$ in the kernel of $x$ and the usual argument shows that $\{y^kv:k\ge 0\}$ linearly spans a finite-dimensional subrepresentation. In this way, we see that every nonzero element $v$ in a locally binilpotent module is contained in a finite-dimensional submodule, since we can prove it when $v$ is an eigenvector of $h$.
So my conjecture implies both conjectures.
The particular case $n=2$ shows the particular case: if $x^2=0$ then $V$ is a sum of irreducible representations of dimension $\le 2$. [Actually the argument works outside characteristic 2,3: indeed since characteristic 2 is excluded, we have $h(h-1)(h+1)=0$ which gives an eigenspace decomposition $V=V_{-1}\oplus V_0\oplus V_1$ for $h$, and $x$ maps $V_1\to V_0\to V_{-1}\to V_{-2}=0$ (that $V_{-2}=0$ uses that characteristic is neither 2 nor 3), and similarly for $y$, etc.]
Added: the conjecture holds for $n=3$.
Write $c=(h+1)^2+4yx$, the Casimir element. Then one gets easily $$q_1:=64x^3y^3=(c-(h-5)^2)(c-(h-3)^2)(c-(h-1)^2);$$ $$q_2:=64yx^3y^2=(c-(h-3)^2)(c-(h-1)^2)(c-(h+1)^2);$$ $$q_3:=64y^2x^3y=(c-(h-1)^2)(c-(h+1)^2)(c-(h+3)^2);$$ $$q_4:=64y^3x^3=(c-(h+1)^2)(c-(h+3)^2)(c-(h+5)^2).$$
Then, for $a(t)=(t+1)(t+2)$ and $b(t)=-3(t+2)(t-1)$, one computes:
$$a(h)q_1+b(h)q_2-b(-h)q_3-a(-h)q_4=8h(h^2-1)(h^2-4)(h^2-3).$$
(The point was to find suitable such $a,b$ so as to eliminate all terms in $c$.)
This looks strange: one does not expect this $h^2-3$ term. But actually it implies that $h$ is diagonalizable (in the quotient of the enveloping algebra by the 2-sided ideal generated by $x^3$), then since $x,y$ maps the $\sqrt{3}$-eigenvalue eigenspace to the $\sqrt{3}\pm 2$ eigenspace, we deduce that the latter is zero.
Note that in finite characteristic, in addition to the expected requirement that the characteristic is not $2,3,5$, we also have to discard characteristic $13$, because in characteristic 13, one has $\sqrt{3}-2=2$. It means that a formula to write $P_2(h)$ as an element of the 2-sided ideal generated by $x^3$ (which I didn't try to write down) might have to involve 13 in the denominator.
Proof in the general case. We fix $n$.
For $0\le k\le n$, one has $$q_k:=(2y)^{n-k}(2x)^n(2y)^k=\prod_{i=0}^{n-1}(c-(h+1-2k+2i)^2).$$
We view $q_k$ as an element of $K[c]$, where $K=\mathbf{C}(h)$ (using that the 0-component of the enveloping algebra is isomorphic to the free commutative algebra $\mathbf{C}[h,c]$).
Recall that for any field $K$ and an $(n+1)$-tuple of distinct elements, the function evaluating a polynomial on this tuple, restricts to a linear isomorphism from the space of polynomials of degree $\le n$ onto $K^{n+1}$.
Let $M$ be the square matrix with indices in $\{0,\dots,n\}$ and entries in $\mathbf{C}(h)$, whose $(i,j)$-entry is the evaluation of $q_j$ at $(h-1-2i)^2$. Then $q_j(h-1-2i)=0$ if and only if $i<j$. So this matrix is lower triangular and invertible.
Hence we can write $1$ as a linear combination of the $q_k=q_k(h,c)$. Clearing denominators and common factors, we can thus write $$P(h)=\sum_{k=0}^nQ_k(h)q_k(h,c),$$ where the family of polynomials in $\mathbf{C}[t]$ $(P,Q_0,\dots,Q_n)$ has gcd equal to 1.
It follows that $P(h)$ belongs to the 2-sided ideal generated by $x^n$. Hence, if $V$ is a representation on which $x^n$ acts by zero, we have $P(h)=0$ and hence we can split a (finite) sum of characteristic subspaces of $h$. It immediately follows that $y$ is nilpotent, and that $h$ acts diagonalizably with eigenvalues in $\{-n+1,\dots,n-1\}$ and $y^n=0$ follows by routine arguments.