I've already proven that if $a, b$ are nilpotent and $ab = ba$ then $(a + b)$ is also nilpotent. But is the condition $ab = ba$ necessary? (Aluffi III.1.6.ii)
I've tried proving that, if $a, b$ are nilpotent and $(a + b)$ is nilpotent as well, then $ab = ba$, but without much success.
I've also tried to construct a counter-example where $a, b$ and $(a + b)$ are all nilpotent but $a, b$ do not commute (by playing around with a few matrices of size 2), but every time I managed to get $a, b$ such that $(a + b)$ is nilpotent, then I magically got $ab = ba$, and I don't know many other examples of non-commutative rings.
Also at this point I don't have a good intuition about this problem. So, do $a, b$ necessarily commute?
No. Let $k$ be a field. In $k^{3\times3}$, take$$a=\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\text{ and }b=\begin{bmatrix}0&0&0\\0&0&1\\0&0&0\end{bmatrix}.$$Then $a$, $b$, and $a+b$ are nilpotent. But $ab\neq ba$.