Here is a proof that this is impossible for commutative rings larger than the continuum or finitely generated maximal ideals.  Let $A$ be an uncountable commutative ring and let $I\subset A$ be a countable maximal ideal.  Then $A$ acts on $I$ by multiplication, giving a homomorphim $\alpha:A\to End_A(I)$.  If either $A$ is larger than the continuum or $I$ is finitely generated, $A$ will have larger cardinality than $End_A(I)$.  In either case, we can conclude that the kernel of $\alpha$ is uncountable.

In particular, we can find some $k\in \ker(\alpha)\setminus I$.  Now by maximality of $I$, there is some $a\in A$ and $i\in I$ such that $ak=1-i$.  But then for any $j\in I$, $0=akj=j-ij$.  This implies $i$ is an idempotent generator of $I$.

Unfortunately, I don't see any obvious way to get rid of the extra hypotheses in this argument.  It _is_ possible for an uncountable ring to act faithfully on a countable ideal; consider $A=\mathbb{Q}^\mathbb{N}$ and $I$ the ideal of sequences that are eventually $0$.  Of course, in this case $I$ fails to be maximal.