The abelian rank of a group $G$ is the maximum $n$ such that $G$ contains an isomorphic copy of $\mathbb{Z}^n$. A group has infinite abelian rank if it contains $\mathbb{Z}^n$ for every $n$.
If $G$ is finitely generated and has infinite abelian rank, does it necessarily have exponential growth? I found a paper by James Reid which seems to say that the answer is yes in the case that $G$ is abelian and torsion-free (in this case, $G$ can be written as the sum of two free subgroups, and free groups have exponential growth).
On the other hand, if a group had intermediate growth, theoretically that would be enough to encompass $\mathbb{Z}^n$ for any $n$ without necessarily being exponential. But, I am not sure.
Bartholdi and Erschler have shown that various permutational wreath products $A\wr_X\Gamma$ (where $\Gamma$ is the first Grigorchuk group and $X$ a suitable transitive $\Gamma$-set) have subexponential growth, and this encompasses $A=\mathbf{Z}$. (See e.g. Lemma 5.1 therein — take $A=\mathbf{Z}\times$ something if necessary.)
So a free abelian group of infinite countable rank embeds (actually, as a normal subgroup) into a finitely generated group of intermediate growth.