If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - t^2$ hold for some $s$ and $t$?

Question

My present question is as is in the title:

If $p^k m^2$ is an odd perfect number with special prime $p$, then must $m^2 - p^k = s^2 - t^2$ hold for some $s$ and $t$?

It is known that $m^2 - p^k$ is not a square.

I note the following:

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For an even perfect number $2^{q-1}(2^q - 1)$ with Mersenne prime $2^q - 1$, then we have $$2^{q-1} - (2^q - 1) = 1 - 2^{q - 1}$$ which is a difference of two squares.

Notice that $$2^{(q-1)/2} < 2^q - 1$$ holds unconditionally.

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If indeed, $m^2 - p^k$ can be written in the form $m^2 - p^k = s^2 - t^2$, and in addition, $1 \leq t \leq 2$, then we have $$(m+s)(m-s) = m^2 - s^2 = p^k - t^2 > 0.$$ This will imply that $$(m+s) \mid p^k - t^2$$ from which it would follow that $$m < m+s \leq p^k - t^2 < p^k.$$

Answer 1

The answer is yes. Since $p\equiv 1\pmod{4}$, and $m$ is odd, we have $m^2-p^k\equiv 0\pmod{4}$. The only integers that are not differences of squares are exactly those that are $2\pmod{4}$.