$(^tA)^n v_1 - v_2 = 0$ has at most one solution for integral hyperbolic matrix $A$

Question

I am stuck with this exercise. Can anyone give me some hints on how to proceed?

Let $A$ be an integral hyperbolic matrix with determinant $\pm 1$ and $v_1, v_2 \in \mathbb{R}^2 \setminus \{ (0, 0) \}$

Show that the equation $(A^T)^n v_1 - v_2 = 0$ has at most one solution $n \in \mathbb{Z}$.

Perhaps I'm wrong but shouldn't $v_1, v_2$ be eigenvectors here?