(**Preamble:** [Andy Putman](https://mathoverflow.net/users/317/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.)
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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 $p$.
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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](https://math.stackexchange.com/a/3122247) by anonymous [MSE user FredH](https://math.stackexchange.com/users/82711/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
$$\sigma(p^k)\sigma(m^2) = 2p^k m^2.$$
We know that $\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$. Since we have shown that $m = (p^k + 1)/2$, then we have the equation
$$2(p^{k+1} - 1)\sigma(m^2) = p^k (p - 1)(p^k + 1)^2. \hspace{0.76in} (*)$$
FredH considered the $GCD$ of $p^{k+1} - 1$ with the right-hand side of Equation $(*)$:
$$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$
where FredH used the fact that $\left(p^{k+1} - 1\right) \mid RHS$ and the property that
$$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$
But FredH also noticed that $p^{k+1} - 1 = p(p^k + 1) - (p + 1)$, whence FredH did also find
$$\gcd(p^{k+1} - 1, p^k + 1) = \gcd(p + 1, p^k + 1),$$
which is $p + 1$ because $k$ is odd. Thus,
$$(p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2 = (p - 1)(p + 1)^2.$$
Hence, the inequality
$$p^{k+1} - 1 \leq (p - 1)(p + 1)^2$$
holds.
Since $k \geq 5$, we obtain
$$p^5 < p^{k+1} - 1 \leq (p - 1)(p + 1)^2 < p^4,$$
which is a contradiction.
Hence, we now have the implication
$$p < m \Rightarrow m^2 - p^k \neq \square.$$
In other words, we have the contrapositive
$$m^2 - p^k = \square \Rightarrow m < p.$$
Now, suppose to the contrary that $m^2 - p^k = \square$. This implies that $m < p$. Since $p^k < m^2$, we then have the implication $m < p \Rightarrow k = 1$. Therefore, $k = 1$. But we know (from the considerations above) that
$$m^2 - p^k = \square \iff m = (p^k + 1)/2.$$
Since $k = 1$, we infer that $m = (p + 1)/2$, or in other words, $p = 2m - 1$. From [Acquaah and Konyagin](https://worldscientific.com/doi/10.1142/S1793042112500935)'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](https://www.lirmm.fr/~ochem/opn/opn.pdf)'s lower bound $N > {10}^{1500}$ for the magnitude of an odd perfect number $N$, together with $p^k < m^2$.)
We conclude that $m^2 - p^k \neq \square$.
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Now, here goes the part where I am a bit unsure about its logical tightness, and is also my main question in this post:
> Does the following statement necessarily hold? "Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares."
If so, WLOG we may assume that
$$(m - a)^2 < m^2 - p^k < (m - a + 1)^2$$
for some positive integer $a$. We may likewise assume that $m > a$.
If I am not mistaken, these assumptions will then yield a proof for the inequality
$$m < p^k < 2am$$
except for the (*problematic*) case $a=1$, where we can only derive
$$p^k < 2m - 1.$$
Either way, I think the inequalities can be summarized as
$$p^k < 2am$$
for some positive integer $a < m$.