Find all $n \in\Bbb N$ such that $2^n -1$ is divisible by $3$ and the congruence $4x^2 \equiv -1\bmod \frac{2^n-1}{3})$ has a solution.

Question

Find all $n \in \Bbb N$ such that $2^n -1$ is divisible by $3$ and the congruence $4x^2 \equiv -1 \bmod\frac{2^n-1}{3})$ has a solution.

I haven proven that $n$ must be even in order to satisfy the first equation. On simulating, I found that $n$ must be a power of $2$. However, I am unable to prove or disprove the latter.