Let $R$ be a commutative ring, $a,b\in R$ such that $a^n=0$ and $b^m=0$ $(n,m\in\mathbb{N})$.
I proved that there exist some $k$ such that $(a+b)^k=0$ (using binomial theorem with $k=n+m$) but such derivation is not very enlightening because take $n=3,m=2$ then $(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4=0$ and $4\neq 5$.
So there's something else going on and I can't see what.
The question would be:
What is the minimal $k$ such that $(a+b)^k=0$?
Yeah, $n+m-1$ is the minimum.
You can take the case $R=\mathbb Z[x,y]/\langle x^n,y^m\rangle.$
Then $$(x+y)^{n+m-2}=\binom{n+m-2}{n-1}x^{n-1}y^{m-1}\neq 0$$
Of course, it is possible in some cases for it to be smaller. If $ab=0$ then you can use $k=\max(m,n).$
If $a=-b$ then $n=m$ and $k=1.$
But the smallest $k$ which works for all rings is $n+m-1.$