By Euclid-Euler theorem, $n = 2^{q-1}(2^{q} - 1)$ where $2^{q} - 1$ is prime. Since $\sigma_{p-2}(n)$ is multiplicative and $(2^{q-1},2^{q} - 1) = 1$,
$$ \begin{align} \sigma_{p-2}(n) &= \sigma_{p-2}(2^{q-1})\sigma_{p-2}(2^{q} - 1)\\ &= (1^{p-2} + 2^{p-2} + \cdot\cdot\cdot +2^{(q-1)(p-2)})(1^{p-2} + (2^{q} - 1)^{p-2})\\ &= (1 + 2^{p-2} + \cdot\cdot\cdot +2^{(q-1)(p-2)})(1 + (2^{q} - 1)^{p-2}). \end{align} $$
From the looks of it I can't find a way to subtract $2$ from the expression then subsequently factor out $p$ (that way we reach the desired conclusion). Any ideas on how to go about this?
You only need $d^{p-1}\equiv 1\bmod p$. Then $$n\sum_{d|n}d^{p-2} \equiv \sum_{d| n}n/d \equiv \sum_{d|n}d \equiv 2n\bmod p$$