Let $\mathbb{F}$ be an algebraically closed field of characteristic zero. Consider $\mathfrak{sl}(2,\mathbb{F})$, with standard basis given by \begin{equation} x= \begin{pmatrix} 0&1 \\ 0&0 \end{pmatrix}, y= \begin{pmatrix} 0&0 \\ 1&0 \end{pmatrix}, h= \begin{pmatrix} 1&0 \\ 0&-1 \end{pmatrix}. \end{equation} In the book "Introduction to Lie Algebras and Representation Theory" by J.E. Humphrey, a maximal vector is introduced as follows.
Let $(\phi,V)$ be an arbitrary finite dimensional $\mathfrak{sl}(2,\mathbb{F})$-module. The semisimplicity of $h$ implies that $V$ decomposes as a direct sum of eigenspaces $V_{\lambda}=\{v\in V\,|\,h.v=\lambda v\}$, where $\lambda\in\mathbb{F}$ is an eigenvalue of $\phi(h)$ (but this definition makes sense for arbitrary $\lambda\in\mathbb{F}$, its just non-zero if and only if $\lambda$ is an eigenvalue). There is a lemma that says that $v\in V_\lambda\implies x.v\in V_{\lambda+2}$, which is trivially proven using the commutation relations. Since $V$ is finite dimensional, there exists some $\lambda\in\mathbb{F}$ such that $V_\lambda\neq0$ but $V_{\lambda+2}=0$, and we say that any $v\in V_\lambda$ is a maximal vector of weight $\lambda$.
Subsequently, there is an exercise which asks you to prove that each arbitrary finite dimensional $\mathfrak{sl}(2,\mathbb{F})$-module has a maximal vector.
Consider the subalgebra spanned $h$ and $x$ (this is actually a hint). It is clear from the commutation relations that this subalgebra is solvable, so there exists a common eigenvector for all elements of this subalgebra. In particular $\phi(h)(v)=\lambda_hv$, $\phi(x)(v)=\lambda_xv$, and $V_{\lambda_h}\neq0$. The nilpotency of $x$ implies the nilpotency of $\phi(x)$, so $\lambda_x=0$ and thus $\phi(x)(v)=0$. That is, $v$ is an maximal vector.
However, we can also say this: Using the finite dimensionality we will find an $\lambda \in \mathbb{F}$ such that $V_{\lambda}\neq0$ and $V_{\lambda+2}=0$, and thus a maximal vector. I don't see anything wrong with this argument.
My question is: Is this fact alone enough to ensure the existence of a maximal vector? If not, why not?
It seems to me that the nilpotency of $x$ is important here, yet there seems to be a proof which does not use it. If someone could clarify/comment on this, and perhaps tell me where I went wrong, that would be nice.