Here is an example.  Choose an uncountable algebraically independent set $S\subset \mathbb{R}$, and let $k=\mathbb{Q}(S)$.  For each $x\in S$, choose a sequence $(q_n(x))$ of rationals converging to $x$.  Given any $y\in k$, write $y=f(x_1,\dots,x_m)$ for some rational function $f$ and some $x_i\in S$ and define $q_n(y)=f(q_n(x_1),\dots,q_n(x_m))$.  Note that $q_n(y)$ may not be defined since the denominator of $f$ could vanish, but for fixed $y$ it is defined for all sufficiently large $n$ since $q_n(x_i)$ must converge to $x_i$.

Now let $A$ be the set of all sequences of rationals $(a_n)$ such that for some $y\in k$, $a_n=q_n(y)$ for sufficiently large $n$.  This set has the same cardinality as $S$, and forms a ring under pointwise addition and multiplication.  There is an epimorphism $\varphi:A\to k$ that sends any sequence to its limit.  The kernel of $\varphi$ is a countable maximal ideal of $A$.  Explicitly, $\ker(\varphi)$ is the set of sequences of rationals that are eventually $0$.  This ideal is clearly not generated by an idempotent.

Note that this construction can give an example having any cardinality up to the continuum.  Here is a proof that for commutative rings, there are no example for which either the ring has cardinality larger than the continuum or the maximal ideal is finitely generated.  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 $I$.