Show that $2 ^ n - 1$ divides $3 ^ n -1$ if and only if $n = 1$

Question

My idea was: If $n$ is even, $\gcd (3ⁿ-1; 2ⁿ-1) = \gcd (6ⁿ-3ⁿ + 3ⁿ-1; 2ⁿ-1) = \gcd (6ⁿ-1; 3ⁿ-1) = 2ⁿ-1 $

So $3ⁿ-1 = 2ⁿ-1$ So $3ⁿ = 2ⁿ$, what does $n = 0$ (nonsense pq gets 0/0)

$3ⁿ-1 \equiv 0 \mod 2ⁿ-1$, so $3ⁿ \equiv 1 \mod 2ⁿ-1$, i.e. $3 \equiv 1 \mod 2ⁿ-1$, $2 \equiv 0 \mod 2ⁿ-1$.

Someone help me?