Does $G/\mathbb{Z} \cong \mathbb{Z}$ imply $G \cong \mathbb{Z}^2$? Let $G$ be a group. If $G/\mathbb{Z} \cong \mathbb{Z}$, does it follow that $G \cong \mathbb{Z}^2$?
More generally: for a normal subgroup $H$ of $G$, if $G/H \cong H$, does it follow that $G \cong H\oplus H$?
 A: No. We can form a semidirect product $\mathbb Z\rtimes\mathbb Z$ with the action $\varphi_a:x\mapsto (-1)^ax$. The result is a non-abelian group $G$ with a subgroup $H\cong\mathbb Z$ for which $G/H\cong\mathbb Z$.
For a more detailed description of this group, see this related question.
A: “More generally”: no. If $C_k$ is the cyclic group of order $k$, then $C_4/C_2\cong C_2$, but $C_2\oplus C_2$ is not cyclic, hence not isomorphic to $C_4$.
For $\mathbb{Z}$, again the answer is “no”. For starters, $G$ need not even be abelian: for instance, you can take the group whose underlying set is $\mathbb{Z}\times\mathbb{Z}$, but with group operation given by
$$(a,b)*(c,d) = (a+(-1)^bc, b+d).$$
This is a semidirect product of $\mathbb{Z}$ with itself. The subgroup of all pairs of the form $(a,0)$ is a normal subgroup isomorphic to $\mathbb{Z}$, and the quotient is isomorphic to $\mathbb{Z}$. But the group is not abelian, since $(0,1)*(1,0) = (-1,1)$, whereas $(1,0)*(0,1) = (1,1)\neq (-1,1)$.
What you can say for the special case where the quotient is $\mathbb{Z}$ (or more generally, a free group), is that if $G/N\cong\mathbb{Z}$, then there exists a subgroup $H$ of $G$ that is isomorphic to $\mathbb{Z}$, with $N\cap H=\{e\}$ and $G=NH$; that is, $G\cong N\rtimes\mathbb{Z}$. For your $G$, there are two possibilities: $\mathbb{Z}\oplus\mathbb{Z}$, and the group I described above. But showing both are the only possibilities is a bit more involved (though not terribly hard).
