Let $N = q^k n^2$ be an odd perfect number with special prime $q$, satisfying $$n^2 - q^k = 2^r t$$ where $r \geq 2$ and $\gcd(2,t)=1$.

We could prove that:

(1)$2^r t > 2n$. (We can modestly improve this to $$2^r t > \frac{3373n^2}{3375},$$ since per this answer to a related MSE question, we have the lower bound $\sigma(n^2)/q^k \geq {3^3} \times {5^3} = 3375$, which implies that $$q^k < \frac{2n^2}{3375}.)$$

(2)$n^2 - q^k$ is not a square.

Our question is:

Must it necessarily be the case that $n > \max(2^r, t)$?

That is, can we rule out the following cases?

(A)$t > n > 2^r$

(B)$2^r > n > t$