In free groups, it is easy to show that two commuting elements are powers of some element:
if $x,y$ are in a free group, then $\langle x,y\rangle$ is free, being subgroup of a free group, with rank $1$ or $2$. If it is of rank $2$, then $x,y$ will not commute, a contradiction. Thus, $\langle x,y\rangle$ is free of rank $1$ i.e. cyclic.
In a paper by Lyndon and Morris Newman, the authors say that similar thing happens in ${\rm PSL}(2,\mathbb{Z})$. They say:
An easy consequence of the Kurosh subgroup theorem (or of simple matrix calculation) is that two commuting elements of ${\rm PSL}(2,\mathbb{Z})$ are necessarily powers of the same element of ${\rm PSL}(2,\mathbb{Z})$.
Any hint to prove this? (both by Kurosh theorem and by Matrix multiplication).