The following is a complete proof that $m^2 - p^k$ is not a square, if $p^k m^2$ is an odd perfect number with special prime $p$. (I apologize in advance for the somewhat lengthy post, I merely had to combine several posts together into one for ease of reference later on.)

Assume that the estimate $p < m$ holds. We want to show that the quantity $m^2 - p^k$ is not a square. Notation-wise, we will denote this conclusion by the shorthand $m^2 - p^k \neq \square$. Suppose to the contrary that $m^2 - p^k = s^2$. It follows that $$(m + s)(m - s) = p^k.$$ Since $p$ is prime, we infer that we have the simultaneous equations $$m + s = p^{k-u}$$ and $$m - s = p^u,$$ where $u$ is an integer satisfying $0 \leq u \leq (k-1)/2$. It follows that we have the system $$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$ and $$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1).$$

We claim that $\gcd(s,p)=1$.

Proof: Suppose otherwise. Then $\gcd(s,p) > 1$. By the definition of GCD, we have both $\gcd(s,p) \mid p$ and $\gcd(s,p) \mid s$. Hence, either $\gcd(s,p) = 1$ (which contradicts our assumption) or $\gcd(s,p) = p$, since the only possible factors of the prime $p$ are $1$ and itself. We infer that $p \mid s$. This implies that $p \mid s^2 = m^2 - p^k$, from which we conclude that $p \mid m$. But this contradicts $\gcd(p,m)=1$.

Since $p$ is the special prime, then $p \equiv 1 \pmod 4$ implies that $\gcd(2,p)=1$. Consequently, from the two simultaneous equations $$2s = p^{k-u} - p^u = p^u (p^{k-2u} - 1)$$ and $$2m = p^{k-u} + p^u= p^u (p^{k-2u} + 1)$$ we obtain that $u = 0$. This implies that $$2s = p^k - 1$$ and $$2m = p^k + 1$$ which is equivalent to $$2s + 1 = p^k$$ and $$2m - 1 = p^k$$ or, expressed differently, as $$s = \frac{p^k - 1}{2}$$ and $$m = \frac{p^k + 1}{2}.$$

Notice that the resulting equation $m = (p^k + 1)/2$ implies that $m < p^k$. Under the assumption $p < m$, then we obtain $k > 1$. Since $k \equiv 1 \pmod 4$, then we know that $k \geq 5$. We can now use a proof by anonymous MSE user FredH to show that $m^2 - p^k \neq \square$ (under the assumption $p < m$), as follows:

Since $N = p^k m^2$ is (odd) perfect, then we have the defining equation $$\sigma(N) = 2N,$$ from which it follows that $$\sigma(p^k)\sigma(m^2) = 2p^k m^2,$$ since the divisor sum $\sigma$ is a multiplicative function.

Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$. From Acquaah and Konyagin's results, we have the unconditional estimate $p < m \sqrt{3}$. This implies that $2m - 1 = p < m \sqrt{3}$, from which we infer that $$m(2 - \sqrt{3}) < 1$$ which contradicts the fact that $\omega(m) > 4$. (In fact, we do know that $m > {10}^{375}$, by using Ochem and Rao's lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)

We conclude that $m^2 - p^k \neq \square$. (Note that, in the case of even perfect numbers $M = 2^{q-1}(2^q - 1)$, the quantity $$2^{q-1} - (2^q - 1) = 1 - 2^{q-1}$$ is likewise not a square.)

Now, here goes the part where I am a bit unsure about its logical tightness:

Either way, I think the inequalities can be summarized as $$p^k < 2am$$ for some positive integer $a$.

Here then, is my question for this post (which, I hope, is acceptable to the MathOverflow community):

Does this "proof" for the inequality $p^k < 2am$ hold water? If not, where does the argument break?

(Preamble: Andy Putman asserts, in the comments, that MO policy prohibits "requests to check completeness of proofs". I have therefore trimmed down my original question to the bare essentials. I hope this would already be OK.)

The following is a proof that $m^2 - p^k$ is not a square, if $p^k m^2$ is an odd perfect number with special prime $p$.

Assume that the estimate $p < m$ holds. We want to show that the quantity $m^2 - p^k$ is not a square. Notation-wise, we will denote this conclusion by the shorthand $m^2 - p^k \neq \square$. Suppose to the contrary that $m^2 - p^k = s^2$. This is true if and only if $$2s + 1 = p^k$$ and $$2m - 1 = p^k.$$ This implies that $p < m < p^k$, from which we obtain $k > 1$. Since $k \equiv 1 \pmod 4$, then we know that $k \geq 5$. We can now use a proof by anonymous MSE user FredH to show that $m^2 - p^k \neq \square$ (under the assumption $p < m$), as follows:

Since $N = p^k m^2$ is (odd) perfect, then we have the defining equation $$\sigma(N) = 2N,$$ from which it follows that $$\sigma(p^k)\sigma(m^2) = 2p^k m^2.$$

Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$. From Acquaah and Konyagin's results, we have the unconditional estimate $p < m \sqrt{3}$. This implies that $2m - 1 = p < m \sqrt{3}$, from which we infer that $$m(2 - \sqrt{3}) < 1$$ which contradicts the fact that $\omega(m) > 4$. (In fact, we do know that $m > {10}^{375}$, by using Ochem and Rao's lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)

We conclude that $m^2 - p^k \neq \square$.

Now, here goes the part where I am a bit unsure about its logical tightness, and is also my main question in this post:

Either way, I think the inequalities can be summarized as $$p^k < 2am$$ for some positive integer $a < m$.

Either way, I think the inequalities can be summarized as $$p^k < 2am.$$

Either way, I think the inequalities can be summarized as $$p^k < 2am$$ for some positive integer $a$.