If $x$ is simple, then this is true. Here is an outline of the proof.
As $x$ and $y$ are freely homotopic, $y=x^g$ for some $g$. As $x$ is simple, $A_x$ and $A_y$ are disjoint.
Consider the Theorem 7.38.3 of Beardon's "Geometry of discrete group." Observe that $g$ is the common perpendicular, therefore $\epsilon =+1$.
This implies if length of $x$ increases, so does length of $x*y$ (in Teichnuller space).
Take right twists along the curve $z$. This will increase the length of $x$ (as $i(x,z)\neq 0$) and hence must increase the length of $x*y$ which implies $i(x*y,z)\neq=0$.
As long as $A_x$ and $A_y$ are disjoint, this proof will work. The other case is when $A_x\cap A_y\neq \emptyset.$ Then $x$ is not simple and their product formula is given by Theorem 7.38.3 of Beardon's "Geometry of discrete group," where $v_2$ will be a lift of the self-intersection point. Now there are two cases:
If $cos\theta$ is positive at $v_2$ then the above arguments hold true.
If $cos\theta$ is negative, it is the same question as thismentioned by Ian Agol for the following reason. Take the self intersection point and consider the two branches of the curves starting and ending at this intersection point. Name them $x$ and $y$. If $y$ is a power of $x$ then you are done by your observation. If not then you get the case $n=1$ of the above question.
