In what follows, we will denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. A positive integer $y$ is said to be perfect when $\sigma(y)=2y$. (Equivalently, $y$ is perfect when $I(y)=2$.)
Let $n = p^k m^2$ be an odd perfect number with special prime $p$, satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. It is fairly obvious that $m^2 - p^k \equiv 0 \pmod 4$.
By MSE user FredH's proof in this paper, we know that $m^2 - p^k$ is not a square. Hence, we may write $$m^2 - p^k = 2^r t$$ for some integer $r \geq 2$ and where $\gcd(2,t)=1$.
Here is my question:
Can we write $r = q - 1$ and $t = 2^q - 1$, where $2^q - 1$ is prime?
That is, if $p^k m^2$ is an odd perfect number, can $m^2 - p^k$ be an even perfect number?
MY ATTEMPT
Assume to the contrary that we can write $m^2 - p^k$ as an even perfect number.
It follows from the considerations above that $m^2 - p^k \neq 6$, since the LHS is congruent to $0$ modulo $4$.
Thus, $r = q - 1$ is even. I also know that $$2^{q-1} = 2^r < t = 2^q - 1.$$
I also know that it would follow from (the assumption) $$I(2^r t) = I(m^2 - p^k) = I(p^k m^2) = 2$$ that $m^2 - p^k$ is a friend of the odd perfect number $p^k m^2$.
Alas, this is where I get stuck.