From this question, we know that subgroups of the rationals is either generated by one element or infinitely generated. The former case is easy to classify: $\left\langle \frac p q \right\rangle$ with $p,q \in \Bbb Z$ and $q \ne 0$.
(Note that if $p=0$, then the subgroup generated is trivial.)
Now, my conjecture is that all infinitely generated proper subgroups of $\Bbb Q$ is of the form $\Bbb Z\left[\frac 1q\right]$ with $q \in \Bbb Z \setminus \{0\}$, i.e. polynomials in $\frac 1q$ with integer coefficients, i.e. $\frac n{q^m}$ with $n \in \Bbb Z$ and $m \in \Bbb N$.
Is my conjecture correct?
This is not correct. Consider for example the subgroup generated by the set $\{p^{-1} \colon p \text{ prime number} \}$.