Let $A$ be a commutative ring, $I \subsetneq A$ a proper ideal of $A$ and $a \in A$ such that $a^k \neq 0$ for all integer $k > 0$. Then there exists an ideal $J$ of $A$ that is maximal satisfying $I\subseteq J$ and $a \notin J$.
A corollary of this statement says that any commutative ring $R$ has a maximal ideal by applying the previous statement with $I=(0)$. But maybe such an $a$ they are talking about does exist : what if every element of $R$ is nilpotent ($nil(R)=R$), i.e. $\forall a \in R, \exists k > 0$ such that $a^k = 0$ ? Since $1$ is nilpotent if and only if (I think) $R=0$ we should hence suppose that this $R\neq 0$ for the corollary to be true right ?
You're exactly right! If $1$ is nilpotent, then $1 = 0$ in $R$. After all, it's easy to show by induction that $1^k = 1$ for all $k$, and then if $1$ is nilpotent then $1 = 1^k = 0$.
I think the first statement is still OK as written, because the zero ring has no proper ideals. For the corollary, I agree that it's false. As you've noticed, if $R$ is the zero ring, it has no proper ideals, thus no maximal (proper) ideals.
I hope this helps ^_^