$V$ is a vector space, $g: V \to V$ an isomorphism and $h: V \to V$ a nilpotent linear map. Also, $g$ and $h$ commute. This implies that $g+h$ is an isomorphism.
Well, $h$ is nilpotent so $h^n=0$. But how to Show that $g+h$ is an isomorphism? I didn't find a way to prove it. I assumed that $g=Id_V$ but i don't know if this works and how to continue.
Take first the case $g = 1$ (the identity map). Say $h^n = 0$. We have: $$(1+h)(1-h + h^2 - \cdots + h^{n-1}) = 1 + h^n = 1$$
Hence $1+h$ is an isomorphism.
Now let $g$ be any isomorphism. We have $g+h = g(1 + g^{-1}h)$. We have that $g^{-1}h$ is nilpotent: say $h^n = 0$; as $g$ and $h$ commute, so do $g^{-1}$ and $h$, so $(g^{-1}h)^n = (g^{-1})^n h^n = 0$. Hence the result follows by above and the fact that the product of isomorphisms is an isomorphism.