On the conjectured inequality $q > k$, where $q^k n^2$ is an odd perfect number with special prime $q$
Solution 1:
On OP's request, I am converting my comment into an answer.
FYI, one can get $\sigma(q^k)\geqslant k+5$ which is better than $\sigma(q^k)\gt k+1$ since $$\dfrac{\partial}{\partial k}\bigg(\sigma(q^k)-k-1\bigg)\gt 0$$ and $$\sigma(q^k)-k-1\geqslant (q+1)-2\geqslant 4.$$