= Solution
The maximum is
$$
\boxed{2}.
$$
The product of two fields $K\times L$ attains it: its only nonzero proper ideals are $K\times0$ and $0\times L$, and both are <maximal ideals>.
For the upper bound, suppose $\mathfrak m_1$ and $\mathfrak m_2$ are distinct maximal ideals of $R$. Their intersection cannot be nonzero, since a nonzero proper ideal is maximal and cannot be properly contained in either of two distinct maximal ideals. Hence $\mathfrak m_1\cap\mathfrak m_2=0$. Also $\mathfrak m_1+\mathfrak m_2=R$, so the <Chinese remainder theorem> gives
$$
R\cong R/\mathfrak m_1\times R/\mathfrak m_2.
$$
Both factors are fields, and this product has exactly two maximal ideals. Thus a third maximal ideal is impossible.
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