Do there exist non-PIDs in which every countably generated ideal is principal?

No such ring exists.

Suppose otherwise. Let $I$ be a non-principal ideal, generated by a collection of elements $f_\alpha$ indexed by the set of ordinals $\alpha<\gamma$ for some $\gamma$. Consider the set $S$ of ordinals $\beta$ with the property that the ideal generated by $f_\alpha$ with $\alpha<\beta$ is not equal to the ideal generated by $f_\alpha$ with $\alpha\leq \beta$.

$I$ is generated by the $f_\beta$ with $\beta \in S$, so if $S$ is finite, then $I$ is finitely generated and thus is principal.

On the other hand, if $S$ is infinite, then take a countable subset $T= \{\beta_1<\beta_2<\dots\}$ of $S$. If the ideal generated by the corresponding set of $f_\beta$'s were principal, its generator would have to be in some $\langle f_{\beta_k} \mid k\leq i \rangle$ for some $i$ (since any element of $\langle f_{\beta}\mid \beta \in T\rangle$ is a finite combination of $f_\beta$'s and therefore lies in some such ideal). Now no $\beta_j$ with $j>i$ could be in $T$.

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The same argument shows that all rings for which any countably generated ideal is finitely generated, have all their ideals finitely generated.

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Corrected thanks to David's questions.

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 ω₁, in the power set P(ω₁), which is a Boolean algebra and hence a Boolean ring. That is, I consists of those subsets of ω₁ that are disjoint from a closed unbounded subset of ω₁. It is an elementary set-theoretic fact that the intersection of any countably many closed unbounded subsets of ω₁ 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.

Sorry to dig up an old question, but in case anybody else randomly lands here, here's a quick side note about a way that this can be generalized.

Theorem: If every countably generated ideal of a ring $R$ is finitely generated, then $R$ is Noetherian. Hence, if $n < \infty$ and every countably generated ideal is $n$-generated, then every ideal is $n$-generated.

Proof: By contrapositive. If $R$ is not Noetherian, then we can make an infinite properly ascending chain $I_1 \subsetneq I_2 \subsetneq \cdots$ of finitely generated ideals. The union of this chain is a countably generated ideal, and it cannot be finitely generated, because that would cause the chain to terminate at some point.