@Carl-FredrikNybergBrodda Whoops - fixed it! Thanks.

I do not understand the sentence "every finite group $G$ of exponent $3$ such as $3$ does not divide $o(G)$". If $G$ is a finite group of exponent $3$ then it has order $3^n$, and so $3$ divides $o(G)$, so no groups satisfy this condition. Right? (I realise this is over 10 years old, so sorry for dredging it up!)

@AchimKrause I was meaning trying to adapt the general proof of the case when $G/\langle x\rangle$ is finite to when $G/\langle x\rangle$ is torsion. It is a standard fact that if $G$ is torsion-free, $\langle x\rangle$ infinite cyclic and $G/\langle x\rangle$ is finite and then $G$ must be infinite cyclic - so can the proof be adapted?

@AchimKrause Yes, I was getting tangled up with groups of this form too. I tried to mimic the proof that if $G$ is torsion-free and $G/\langle x\rangle$ is finite then $G$ is cyclic, and the idea here is to prove that $G$ splits. So for example if $H^2(G/\langle x\rangle, \mathbb{Z})=0$ then we'd be done, but I cannot see why this would be $0$.

@MikaeldelaSalle I meant to exclude that possibility! I'll edit the question to rule it out.