Let $d > 0$ be square-free. Let $\epsilon = x_0 + y_0 \sqrt{d}$ be the minimal solution to the Pell's equation $x ^ 2 - d y ^ 2 = 1$. Let $x + y \sqrt{d} = \epsilon ^ l, l \geq 1$ be a solution.
Question: If all the prime divisors of $x$ divides $x_0$, does it follow that $l = 1$, i.e. $x + y \sqrt{d} = x_0 + y_0 \sqrt{d}$ (or in other words, $x + y \sqrt{d}$ is the minimal solution) ?