When trying to show that $I+J$ is nilpotent, whenever $I,J$ are nilpotent ideals of a Lie algebra $L$, I did it brute force: By induction we can show the following: An element of $(I+J)^{2N}$ is a sum of elements of the form $[x_1,[x_2,[\ldots,[x_{2N},x_{2N+1}]]\ldots]$, with $x_i \in I \cup J$. So at least $N$ of the $x_i$ lie in $I$ or in $J$. Using induction again, one can show that this product then lies in $I^N$ resp. $J^N$. Hence, if we choose $N$ large, it follows that $(I+J)^N = 0$.
I was wondering if there is some more elegant proof. Is it maybe easier to use Engel's theorem ?
We have $(I+J)^{2m}\subseteq I^m+J^m$ for all $m\ge 1$, hence $(I+J)^{2m}=0$ for $m$ large enough. I find this proof very natural, but indeed there is another proof using Engel's theorem and another two lemmas from representation theory; so it is perhaps more elegant, but it is also more complicated. This depends on your taste, too. For the alternative proof see Lemma $5.3.3$, Lemma $5.3.4$ and Lemma $5.3.5$ here.