On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II (Preamble: We have asked this same question in MSE two weeks ago, without getting any answers.  We have therefore cross-posted it to MO, hoping that it gets answered here.)
The topic of odd perfect numbers likely needs no introduction.
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
If $n$ is odd and $\sigma(n)=2n$, then we call $n$ an odd perfect number.  Euler proved that a hypothetical odd perfect number must necessarily have the form $n = p^k m^2$ where $p$ is the special prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds.  Dris conjectured that the inequality $p^k < m$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $p < m$.
Now, recent evidence suggests that $p^k < m$ may in fact be false.
THE ARGUMENT
Let $n = p^k m^2$ be an odd perfect number with special prime $p$.
Since $p \equiv k \equiv 1 \pmod 4$ and $m$ is odd, then $m^2 - p^k \equiv 0 \pmod 4$.  Moreover, $m^2 - p^k$ is not a square (Dris and San Diego (2020)).
This implies that we may write
$$m^2 - p^k = 2^r t$$
where $2^r \neq t$, $r \geq 2$, and $\gcd(2,t)=1$.
It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:
$$\text{Case (1):  } m > t > 2^r$$
$$\text{Case (2):  } m > 2^r > t$$
$$\text{Case (3):  } t > m > 2^r$$
$$\text{Case (4):  } 2^r > m > t$$
$$\text{Case (5):  } t > 2^r > m$$
$$\text{Case (6):  } 2^r > t > m$$

We can easily rule out Case (5) and Case (6), as follows:
Under Case (5), we have $m < t$ and $m < 2^r$, which implies that $m^2 < 2^r t$.  This gives
$$5 \leq p^k = m^2 - 2^r t < 0,$$
which is a contradiction.
Under Case (6), we have $m < 2^r$ and $m < t$, which implies that $m^2 < 2^r t$.  This gives
$$5 \leq p^k = m^2 - 2^r t < 0,$$
which is a contradiction.

Under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds, as follows:
Under Case (1), we have:
$$(m - t)(m + 2^r) > 0$$
$$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case (2), we have:
$$(m - 2^r)(m + t) > 0$$
$$p^k = m^2 - 2^r t > m(2^r - t) = m\left|2^r - t\right|.$$

So we are now left with Case (3) and Case (4).
Under Case (3), we have:
$$(m + 2^r)(m - t) < 0$$
$$p^k = m^2 - 2^r t < m(t - 2^r) = m\left|2^r - t\right|.$$
Under Case (4), we have:
$$(m - 2^r)(m + t) < 0$$
$$p^k = m^2 - 2^r t < m(2^r - t) = m\left|2^r - t\right|.$$
Note that, under Case (3) and Case (4), we actually have
$$\min(2^r,t) < m < \max(2^r,t).$$
But the condition $\left|2^r - t\right|=1$ is sufficient for $p^k < m$ to hold.
Our inquiry is:

QUESTION: Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case (3) and Case (4)?

Note that the condition $\left|2^r - t\right|=1$ contradicts the inequality
$$\min(2^r,t) < m < \max(2^r,t),$$
under the remaining Case (3) and Case (4), and the fact that $m$ is an integer.
 A: You wrote

$$\implies (a + 1)p^k + b(a + 1) = p^k + b + ap^k + ab = p^k + c$$Comparing coefficients, we obtain$$a + 1 = 1 \text{ and } b(a + 1) = c$$

I don't think that this is correct since it is not true that
$$(a + 1)p^k + b(a + 1)  = p^k + c\implies a+1=1$$
For example, if
$$a=2,\quad b=2,\quad c=16,\quad p=5,\quad k=1$$
then
$$(a + 1)p^k + b(a + 1) = p^k + c$$
holds.
A: 
QUESTION: Is the condition $\left|2^r - t\right|=1$ also necessary for $p^k < m$ to hold, under Case (3) and Case (4)?

The answer is no.
It is true that $p^k\lt m\implies |2^r-t|\not=1$.
Proof :
Since you ruled out (5) and (6), and showed that, under (1) or (2), $m < p^k$ holds, one can say that it is true that $p^k\lt m\implies$ (3) or (4).
Also, since, under (3) or (4), $2^r$ and $t$ cannot be consecutive integers, one can say that $|2^r-t|\not=1$.
As a result, one can say that
$$p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1.\quad\blacksquare$$
A: Edit: October 30, 2021 - 2:56 PM Manila time This solution is flawed, as detailed in this other answer by mathlove.

Consider the remaining Case (3) and Case (4) in what follows.

Let $p^k < m$, and assume that $\left|2^r - t\right| > 1$.  Then
$$\left|2^r - t\right| = 1 + a,$$
where $a > 0$.
So we have:
$$m = p^k + b,$$
where $b \geq 2$,
$$p^k < m \left|2^r - t\right| \implies m \left|2^r - t\right| = p^k + c,$$
where $c > 0$,
$$\implies (p^k + b)(1 + a) = p^k + c$$
$$\implies (a + 1)p^k + b(a + 1) = p^k + b + ap^k + ab = p^k + c$$
Comparing coefficients, we obtain
$$a + 1 = 1 \text{ and } b(a + 1) = c$$
so that we get
$$a = 0 \text{ and } b = c.$$
This contradicts $a > 0$.

Hence, $p^k < m$ is equivalent to $\left|2^r - t\right| = 1$, which contradicts
$$\min(2^r,t) < m < \max(2^r,t),$$
thus finally yielding a complete proof for the inequality $m < p^k$.
