if $a,b$ are nilpotent elements of a commutative ring $R$, show that $a+b$ is also nilpotent

You can use the binomial theorem without working out the binomial coefficients if it makes it easier for you. Take $(a+b)^{m+n} = \sum_{i=0}^{m+n}c_i a^{i}b^{m+n-i}$ where $c_i$ is what the coefficients should be. Note for each $i \le n$, $b^{m+n-i} =0$ and for $i \ge n$, $a^i = 0$. Therefore each element of this sum is $0$.

Note however that this is false in a general ring, i.e. the sum of nilpotent elements needs not be nilpotent in general. Take as your ring $M_{2\times 2}(F)$ with $a = \left[\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array}\right]$ and $b = \left[\begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array}\right]$. Then $a^2 = b^2 = \left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right]$, but $(a+b)^2 = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]$ which is certainly not nilpotent.

I'm not sure how to avoid the binomial theorem, but here is a way to prove the proposition.

$$(a+b)^{n+m}=\sum_{k=0}^{n+m}{n+m\choose k} a^kb^{n+m-k}=\sum_{k=0}^{n-1}{m+n\choose k}a^kb^{m+n-k}.$$

The last equality follows because $a^k=0$ for $n\le k\le m+n$. So, $k<n$ for the terms that remain, which implies $m+n-k>m$. So, $b^{m+n-k}=0$ and these terms vanish also.

$ (a\!+\!b)^{m+n}\! =\! \overbrace{\sum\, c_{ij}\,a^i b^j}^{\textstyle i\!+\!j \color{#0a0}{= m\!+\!n}}\!\Rightarrow\ \begin{align}\overbrace{i\ge\, n^{\phantom{I}}\!\!\!}^{\Large a^i\,=\ 0}\\[.1em] \color{#c00}{{\rm or}}\,\ \ \underbrace{j\ge m}_{\Large b^j\,=\ 0}\end{align}\,\,$ $\ (\color{#c00}{\rm else}\, $ $ \begin{align}i<m\\ j<n\,\end{align}$ $\Rightarrow$ $\, i\!+\!j \color{#0a0}{< m\!+\!n})$