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.