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 $I$.