Suppose that $G$ is an infinite group, $a,b\in G$ satisfy $G=\langle a,b\rangle$ and $a^n=b^n=e$ for some $n\in\mathbb{N}^*$, where $e$ is the identity of $G$. Must $G$ contain infinitely many $x$ such that $x^n=e$?
The result is almost obvious for $n=2$: we have $((ab)^ra)^2=(b(ab)^s)^2=e$ for all $r,s\in\mathbb{N}$. If $x^2=e$ has only finitely many solutions, then $(ab)^ra=(ab)^{r'}a$ for some $r\neq r'$, or $b(ab)^s=b(ab)^{s'}$ for some $s\neq s'$, or $(ab)^ra=b(ab)^s$ for some $r,s$. In either case, there is some $m$ such that $(ab)^m=e$ ($e=r'-r$ in the first case, $s'-s$ in the second case and $r+s+1$ in the third case), and it turns out that $G$ is a subgroup of $D_{2m}$, the dihedral group of order $2m$, which is a finite group.
What can be said for general $n$? Any help appreciated.
Edit: Since the result has been proven to be true, there is a consequence that may appear to be interesting: suppose that every finite subgroup of $G$ is cyclic, then for every $n\in\mathbb{N}^*$, $x^n=e$ has either at most $n$ solutions or infinitely many solutions.
Prove: Fix $n$, suppose that $x^n=1$ has only finitely many solutions. Let $S$ be the set of solutions. If there exists $a,b\in S$ such that $\langle a,b\rangle$ is infinite, then $\langle a,b\rangle$ contains infinitely many $x$ such that $x^n=e$, a contradiction; so the subgroup generated by every two elements in $S$ is finite (and thus cyclic), which means that every two elements in $S$ commute, so $\langle S\rangle$ is finite (and thus cyclic, actually equal to $S$), which means that $x^n=1$ has at most $n$ solutions.
Of course the converse is not true, as shown by the example $D_\infty\times C_{2m}$, where $D_\infty$ is the infinite dihedral group ($x^n=1$ has at most $n$ solutions for odd $n$ and infinitely many solutions for even $n$, but the group contains infinitely many subgroups isomorphic to $C_2\times C_{2m}$ which is not cyclic).
Edit 2: The result above is already contained by the comment of Mikko Korhonen to the answer.
To provide an answer, here is a summary of the comments.
The answer to the question is yes, there are infinitely many elements in $G$ with $x^n=1$.
For suppose not. Then $a$ and $b$ have finitely many conjugates, so their centralizers $C_G(a)$ and $C_G(b)$ both have finite index in $G$, and hence so does their intersection $C_G(a) \cap C_G(b)$. But $C_G(a) \cap C_G(b) = C_G(\langle a,b \rangle) = C_G(G) = Z(G)$, so $|G:Z(G)|$ is finite.
Now, by a well-known result of Schur (the proof is not difficult provided that you are familiar with the transfer homomorphism), $|G:Z(G)|$ finite implies that $[G,G]$ is finite. Then, since the images of $a$ and $b$ have finite order in the abelian group $G/[G,G]$ which they generate, $G/[G,G]$ is finite and hence $G$ is finite, contrary to the assumption.