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This is a followup to my question here.

Assume $A$ is a ring, and $B \subset A$ is a multiplicative subset. The prime ideals of the localization $B^{-1}A$ can be identified with those prime ideals of $A$ containing no elements of $B$. If $A$ is finitely generated over $k = \overline{k}$, how do I see that the maximal ideals of $B^{-1}A$ can be identified with those maximal ideals of $A$ containing no elements of $B$?

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    The correct statement is: maximal ideals in the localisation can be identified with ideals in $A$ that are maximal among those who contain no elements of $B$. – 2017-02-22

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It's not necessarily possible to do so!

Let $k$ be an algebraically closed field, $A = k[x,y]$, and let $B$ be multiplicative subset of all nonzero polynomials in $x$.

Every maximal ideal of $A$ is of the form $(x-a, y-b)$, and thus every maximal ideal of $A$ contains an element of $b$.

However, $B^{-1} A = k(x)[y]$, and has plenty of maximal ideals.