Not only the answer is yes (as indicated in Adam's answer), but there's a full picture:

$G$ is a compact abelian group with a homomorphism with dense image $\mathbf{Z}\to G$. "With dense image" means an epimorphism in the category of locally compact abelian groups. This means that its Pontryagin dual $\hat{G}$ is a discrete abelian group comes with a monomorphism into the circle group $\mathbf{R}/\mathbf{Z}$, where monomorphism simply means injective homomorphism. In other words, $G$ is obtained by considering the Pontryagin dual of any subgroup of the circle group endowed with the discrete topology. That $G$ is metrizable means that $\hat{G}$ is countable.

Pontryagin duality yields a bijection between the set of closed subgroups of $G$ and the set of subgroups of $\hat{G}$.

Now discrete abelian groups with countably many subgroups were characterized by Boyer (1956) and are pretty rare: these are abelian groups $H$ with a finitely generated subgroup $A$ such that $H/A$ is isomorphic to $\bigoplus_{p\in I}C(p^\infty)$, where $I$ is a finite set of primes (no multiplicity allowed). Here $C(p^\infty)$ is the Prüfer (quasi-cyclic) group $\mathbf{Z}[1/p]/\mathbf{Z}$.

So it's enough to consider any group not of this form, and embeddable in the circle group. For a subgroup $H$ of the circle group, there are three ways to fail to satisfy the above criterion:

- to have infinite $\mathbf{Q}$-rank: for instance, one considers free $\mathbf{Z}$-module of infinite countable rank: this yields $G$ to be the infinite-dimensional torus as in Adam's answer;
- to have $p$ torsion for infinitely many primes $p$. For instance, for some infinite set $J$ of primes $H=\bigoplus_{p\in J}\mathbf{Z}/p\mathbf{Z}$, and then $G=\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$.
- to have twice the same prime occurring in the Prüfer part. This holds if $H\simeq\mathbf{Z}[1/p]^2$, so $G=\hat{H}\simeq S^2$ where $S$ (the Pontryagin dual of $\mathbf{Z}[1/p]$) is called "solenoid" (unlike the last two previous examples it's not straightforward here that $G$ has uncountably many closed subgroups, since $S$ itself has only countably many ones, but still it's true).