= Solution
Let $\mathfrak m$ be maximal in $R/I$ and let $\mathfrak n\subseteq R$ be its contraction. The ideal $\mathfrak n$ is maximal and contains $I$. It is disjoint from $S=1+I$: if $1+i\in\mathfrak n$ with $i\in I\subseteq\mathfrak n$, then $1\in\mathfrak n$. The <prime ideal correspondence for localization> therefore defines the proper ideal $S^{-1}\mathfrak n$, and
$$
S^{-1}R/S^{-1}\mathfrak n\cong R/\mathfrak n
$$
is a field. Thus $f(\mathfrak m)=S^{-1}\mathfrak n$ is maximal.
Conversely, let $\mathfrak q$ be the contraction to $R$ of a maximal ideal of $S^{-1}R$. Then $\mathfrak q$ is maximal among ideals disjoint from $S$. Since $\mathfrak q+I$ is disjoint from $S$—otherwise an equation $q+i=1+j$ would put $1+(j-i)\in\mathfrak q\cap S$—maximality gives $I\subseteq\mathfrak q$. If a proper ideal strictly contained $\mathfrak q$ in a maximal ideal $\mathfrak n$ of $R$, then $\mathfrak n$ would also contain $I$ and hence remain disjoint from $S$, a contradiction. Thus $\mathfrak q$ is maximal and contains $I$.
The two standard extension-contraction bijections, first for $R\to R/I$ and then for $R\to S^{-1}R$, now give inverse maps
$$
\boxed{\operatorname{MaxSpec}(R/I)\cong\operatorname{MaxSpec}(S^{-1}R)}.
$$
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