For given $x,y \in \mathfrak{g}, [x,y], [x,[x,y]] \neq 0$, does there exist a $z \neq 0 \in \mathfrak{g}$ such that
$$\operatorname{ad}^2_xy = \operatorname{ad}_zy$$
For simplicity, let's think of the case of $\mathfrak{g} = End(V)$, with the commutator $\operatorname{ad}_xy = xy-yx$
I suspect the answer is no, but I can't think of "proper" reason other than computing the LHS and seeing terms like $xyx$ that a priori will dissallow writing a $zy$.
Edit: as pointed out by user1551 there are certain matrices that this will never work, but it's not a done deal, when and why does this happen? Also note that if we work with the Lie algebra of vectors in $\mathbb{R}^3$ with the cross product, my question becomes when can the BAC - CAB triple product identity be expressed as a cross product. i.e. solving:
$$x(x \cdot y) - y\|x\|^2 = z \times y$$
Thus, a partial proof for the $\mathfrak{s}\mathfrak{o}(3)$ case is if $x \perp y$ the resulting equation reads:
$$-\|x\|^2 y = z \times y$$
Which can have no nontrivial solution, as geometrically we are asking $y$ to be orthogonal to a scalar multiple of itself. This is what happens in user1551's example too, as the matrices chosen are orthogonal under the trace pairing, which is the Killing form for the Lie algebra $End(V)$, thus to improve my question, suppose the case that $x \not\perp y$, or when the Killing form is degenerate and we can't define such notions of orthogonality.
Edit2: see the comments for a very slick proof by user1551 that the solution to this problem is negative for $\mathfrak{g}=\mathfrak{s}\mathfrak{o}(3)$
All nontrivial split semisimple Lie algebras (and hence all the ones which would have such as quotient) contain counterexamples.
In $\mathfrak g:= \mathfrak{gl}_n(k) \simeq (M_n(k), [\cdot,\cdot])$ for $n\ge 2$, take $x=\pmatrix{0&\dots &1\\\vdots&&\vdots\\ 0&\dots&0}$, $y=\pmatrix{0&\dots &0\\\vdots&&\vdots\\ 1&\dots&0}$. Then $ad_x(ad_x(y))=-x$ but there is no $z\in \mathfrak g$ with $ad_z(y) =-x$.
In fact, if $\mathfrak g$ is any split semisimple Lie algebra with a chosen Cartan subalgebra, corresponding roots $\alpha \in \Phi$ and root spaces $\mathfrak g_{\alpha}$, then for any given root $\alpha$ and bases $e_\alpha$ of $\mathfrak g_\alpha$, take $x=e_\alpha$ and $y=e_{-\alpha}$. Then $ad_x(ad_x(y))$ is a non-trivial element of $\mathfrak g_\alpha$ but there is no element $z$ such that the $\alpha$-component of $ad_z(y)$ would be nonzero.
One can probably generalize this even further to non-split cases. Note also that the two elements are not orthogonal wrt the Killing form.
(And of course, over there are also always non-trivial cases where solutions exist. In fact, if $[x,y] = \lambda y$ for a nonzero $\lambda \in k$, then $ad_z(y) = ad_x(ad_x(y))$ for $z=\lambda x$.)