Since $n$ is an odd pseudoprime it means $2^n\equiv 2(\bmod n)$, hence $2^n - 2 = kn$ for some $k\in \mathbb{Z}$.
Now $m=2^n - 1$ is not prime since $n$ is not prime, so it sufficies to show $2^m\equiv 2(\bmod m)$. Note that $2^{m-1} = 2^{2^n - 2} = 2^{kn}$. Therefore, $2^{m-1} - 1$ is divisible by $2^n - 1=m$ since $n|(m-1)$. Thus, $2^{m-1} \equiv 1(\bmod m) \implies 2^m \equiv 2(\bmod m)$.