It is well known that $2^p-1$ is prime only if $p$ is prime($2^a-1|2^{ab}-1$). Applying the lemma twice, gives the result.

Claim: If $2^x - 1$ is a prime for some $x \in \mathbb Z_{>0}$, then $x$ is a prime. Proof: If not, consider a divisor $d \notin \{1,x\}$ of $x$. Then $2^d -1 \mid 2^x - 1$ while $2^d - 1 \notin \{1,2^x -1\}$, meaning $2^x - 1$ is not a prime, contradiction. $\square$ So our claim gives: $$2^{\color{green}{2^n -1}} - 1 \text{ is prime} ~ \implies ~ 2^{\color{green}{n}} - 1 \text{ is prime} ~ \implies \boxed{n \text{ is prime}}$$