Maximal \(\implies\) prime, but generally not the converse.
Solution
Suppose \(\mm\) is maximal, \(ab\in \mm\), and \(b\not\in \mm\).
Then there is a containment of ideals \(\mm \subsetneq \mm + (b) \implies \mm + (b) = R\).
So \[ 1 = m + rb \implies a = am + r(ab) ,\] but \(am\in \mm\) and \(ab\in \mm \implies a\in \mm\).
The ideal \((0) \subset \ZZ\) is prime since \(\ZZ\) is a domain, but not maximal since \((0) \subsetneq (2) \subsetneq \ZZ\).