Let $G$ be a group (finitely generated) and let $N \triangleleft G$ be virtually solvable such that $G/N$ is virtually infinite cyclic. We also assume that $N$ is finitely generated (this assumption was omitted in the first version)
We aim to show $G$ is virtually solvable (I need a sanity-check for this proof).
I won't fill in the calculations but only show the outline.
Since $N$ is virtually solvable we have the following short exact sequence:
$$1 \longrightarrow N_0 \longrightarrow N \longrightarrow N/N_0 \longrightarrow 1$$
where $N_0$ is solvable and the quotient is finite.
Since $G/N$ is virtually infinite cyclic we have the short exact sequence:
$$1 \longrightarrow H/N \longrightarrow G/N \longrightarrow G/H \longrightarrow 1$$
where $H \triangleleft G$, $H/N$ is infinite cyclic and $G/H$ is finite.
Now we have that the below s.e.s splits because $H/N$ is free:
$$1 \longrightarrow N \longrightarrow H\longrightarrow H/N \longrightarrow 1$$
Therefore, $H \cong N \rtimes H/N $ where the action of the quotient on $N$ is given by conjugation.
Consider this theorem by Hall:
We have that $N$ is finitely generated and normal in $G$, $N_0$ has finite index in $N$ so there is $N_1 \subset N_0$, $N_1$ normal in $G$ and $[N:N_1] < \infty$.
Thus the index, $[N \rtimes H/N:N_1 \rtimes H/N]$, is finite (because $[N:N_1]< \infty$ - is this accurate?).
Using the above theorem again, and with the observation that $N \rtimes H/N$ is finitely generated there is a subgroup $L \subset N_1 \rtimes H/N$ which is normal in $N \rtimes H/N$ and has finite index.
Finally, $N_1 \rtimes H/N$ is solvable because $[N_1 \rtimes H/N, N_1 \rtimes H/N] \subset [N_1,1] \cong N_1$ which is solvable, hence $L$ is solvable and of finite index in $H$, and normal in $G$. Hence $G$ is virtually solvable.
Please correct me if something is off, thanks :)
You can make a few improvements that would help. First, since you are working up to finite index you may as well assume that $G/N$ is cyclic, rather than just virtually cyclic. Second, your $N$ is virtually soluble, so there is a subgroup $N_0$ of $N$, the soluble radical of $N$, and this has finite index in $N$. Furthermore, since it's characteristic in $N$, $N_0\lhd G$, and so we may assume that $N_0=1$ if we are only proving that $G$ is virtually soluble. Hence all you need to show is that finite-by-cyclic is cyclic-by-finite.
To do this last bit, it's fairly easy, and you don't any big theorems, or even freeness of cyclic groups. If $|N|$ is finite and $G/N$ is cylcic, let $x\in G$ be the preimage of a generator for $G/N$. Then $x$ has infinite order and $G=\langle N,x\rangle$, so $G=N\rtimes H$, where $H=\langle x\rangle$. Finally, $x$ normalizes $N$, a finite group, so some power $x^n$ of $x$ must centralize $N$. Thus $\langle x^n\rangle$ is a normal subgroup, and its index is $|N|\times n$.