Let $p$ be prime and let $d\in\mathbb{Z}_{>1}$ dividing $2^p-1$. Show that $d$ is prime or Fermat-pseudoprime with respect to 2.
I'd like to prove this by showing that $p$ divides $d-1$. Because, since we are given that $2^p\equiv1\mod d$, we have by raising the equivalence to a suitable power, that indeed $2^{d-1}\equiv1\mod d$. However, I get stuck at proving that $p$ divides $d-1$.
Let $q$ be a prime factor of $d$, and let $k > 1$ be the order of $2$, mod $q$.
Since $q$ is prime, it follows that $k{\;|\,}(q-1)$.
\begin{align*} \text{Then}\;\;&d{\;|\,}(2^p-1)\\[4pt] \implies\;&q{\;|\,}(2^p-1)\\[4pt] \implies\;&2^p\equiv 1\;(\text{mod}\;q)\\[4pt] \implies\;&k{\,|\,}p&&\text{[since $k$ is the order of $2$, mod $q$]}\\[4pt] \implies\;&k=p&&\text{[since$\;p\;$is prime and$\;k > 1$]}\\[4pt] \implies\;&p{\;|\,}(q-1)&&\text{[since$\;k{\;|\,}(q-1)$]}\\[4pt] \implies\;&q\equiv 1\;(\text{mod}\;p)\\[4pt] \end{align*} Thus, every prime factor of $d$ is congruent to $1$, mod $p$.
It follows that $d\equiv 1\;(\text{mod}\;p)$, or equivalently, $p{\;|\,}(d-1)$.
Then $(2^p-1){\,|\,}(2^{d-1}-1)$, hence from $d{\;|\,}(2^p-1)$, we get $d{\;|\,}(2^{d-1}-1)$, or equivalently, $2^{d-1}\equiv 1\;(\text{mod}\;d)$.
Therefore $d$ is a prime or a Fermat-pseudoprime, base $2$, as was to be shown.