In a nonzero commutative ring, every proper ideal is contained in a maximal ideal

Let $\mathcal P$ be the poset of proper ideals of $R$ containing $I$, ordered by inclusion.

$\mathcal P$ is nonempty, because $I$ is a proper ideal containing $I$.

The empty chain of $\mathcal P$ has an upper bound in $\mathcal P$: every member of $\mathcal P$ is vacuously above all of its members, and $\mathcal P$ is nonempty, so $I$ is such an upper bound.

A nonempty chain $\mathcal C\subseteq\mathcal P$ has an upper bound in $\mathcal P$: $\bigcup\mathcal C$ is an ideal containing $I$, and it is proper, since $1\in\bigcup\mathcal C$ would place $1$ in some member of $\mathcal C$, forcing that member to equal $R$ and contradicting its properness.

Every chain of $\mathcal P$ has an upper bound in $\mathcal P$, and $\mathcal P$ is a nonempty poset, so Zorn's lemma yields a maximal element $M$ of $\mathcal P$.

$M$ is a proper ideal containing $I$ that is maximal among proper ideals of $R$, so $M$ is a maximal ideal containing $I$.