Prove that if ${\frak g}\le{\frak gl}(V)$ and $\mathfrak h\le\mathfrak g$ is a maximal subalgebra, then $\mathfrak h$ is an ideal of codimension $1$

Question

Let ${\frak g}\le{\frak gl}(V)$ be a finite-dimensional subalgebra of nilpotent endomorphisms on some vector space $V$, and let ${\frak h}\le{\frak g}$ be a maximal subalgebra of $\mathfrak g$.

Using Engel's theorem, one can see that this implies that $\mathfrak h$ must be an ideal of $\mathfrak g$ with codimension $1$. A sketch of the argument goes as follows:

1. $\mathfrak h$ acts on $\mathfrak g/\mathfrak h$ via its adjoint: $h.(g+\mathfrak h)\equiv [h,g]+\mathfrak h$. This action is via nilpotent operators (because if $h$ is nilpotent then $\operatorname{ad}(h)$ is also nilpotent);
2. from Engel's theorem, we know that a subalgebra of nilpotent operators acting on a vector space has a common zero eigenvector. This implies here that there is some $x\in\mathfrak g\setminus\mathfrak h$ such that $[\mathfrak h,x]\subseteq\mathfrak h$;
3. If $\mathfrak h$ did not have unit codimension, there would be some $y\notin\operatorname{span}(\{\mathfrak h,x\})$, and thus $\operatorname{span}(\{\mathfrak h,x\})$ would be a proper subalgebra containing $\mathfrak h$. Being $\mathfrak h$ maximal, we conclude that $\mathfrak h$ has unit codimension and is an ideal.

I've seen this argument used to prove Engel's theorem via induction.

Is there a more direct way to prove this without passing through Engel's theorem?