Given a torsion-free hyperbolic group $G$, does there exist a number $n(G)$ such that for any $x,y,z\in G$, $x^n y^n z^n =1$ implies that $x$, $y$, and $z$ commute pairwise?

Some musings/questions... When $G$ is free, the result is true for $n=2$.

Clearly, this does not generalize to hyperbolic groups (e.g. non-orientable surface groups of genus 3).

Somewhat related: given any 3 non-commuting elements $x,y,z\in G$ there is a number $n(x,y,z)$ such that $ < x^n, y^n , z^n \> $ is free. Is this true??..definitely in case of two elements (Gromov).