Given a finite dimensional Lie algebra $L$, suppose that each maximal lie subalgebra of $L$ is an ideal. Suppose the adjoint map, $\operatorname{ad}_y$ is not nilpotent.
Then pick a maximal subalgebra $M$ containing $L_{0,y} := \{x \in L |(\operatorname{ad}_y)^n(x) = 0 \space \text{for some} \space n\in \mathbb{N}\}$
Then $M$ is an ideal by assumption. And $\dim(L/M) = 1$.
But $y \in L_{0,y} \leq M$ and so $[y,x] \in M \quad \forall x\in L$.
So $\operatorname{ad}_y$ acts like $0$ on $L/M$.
$\bf{Question}$ : How can I then deduce that $\operatorname{ad}_y$ has eigenvalue $0$ on $L/L_{0,y}$?