The question is fully settled by Hugh Thomas' anwer, but let me mention this related interesting fact.

**Theorem.** There is a ring R and ideal I on R, such that every countable subset
of I is contained in a principal subideal of I, but I is
not principal.

Proof. Let I be the ideal of nonstationary subsets of
ω_{1}, in the power set P(ω_{1}), which is a Boolean
algebra and hence a Boolean ring. That is, I consists of those subsets of ω_{1} that
are disjoint from a closed unbounded subset of ω_{1}. It is an elementary set-theoretic fact that the intersection of any countably many closed unbounded subsets of ω_{1} is still closed and unbounded, and thus the union of countably many
non-stationary sets remains non-stationary. Thus, every
countable subset of I is contained in a principal subideal
of I. But I is not principal, since the complement of any
singleton is stationary. QED

In the previous example, the ideal I is not maximal. If one assumes the existence of a measurable cardinal (a large cardinal notion), however, then the example can be made with I maximal.

**Theorem.** If there is a measurable cardinal, then there is a
ring R with a maximal ideal I, such that every countable
subset of I is contained in a principal sub-ideal of I, but
I is not principal.

Proof. Let κ be a measurable cardinal, which means that
there is a nonprincipal κ-complete ultrafilter U on the
power set P(κ), which is a Boolean algebra and thus a
Boolean ring. The ideal I dual to U is also κ-complete,
which means that I is closed under unions of size less than
κ. In particular, since kappa is uncountable, this
means that the union of any countably many elements of I
remains in I, and this union set generates a principal
subideal of I containing the given countable set. The ideal I is maximal since U was an ultrafilter. QED

I'm not sure at the moment whether the situation of this last theorem requires a measurable cardinal or not, but I'll think about it.