On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part II 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 rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.
So we are now left with Case (3) and Case (4):
Under both cases left under consideration, we have
$$(m - 2^r)(m - t) < 0$$
$$m^2 + 2^r t < m(2^r + t)$$
$$m^2 + (m^2 - p^k) < m(2^r + t)$$
$$2m^2 < m(2^r + t) + p^k.$$
Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$.  We get
$$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$
which implies, since $m > 0$, that
$$2m < 2^r + t + 1.$$
Here then is our question:

Will it be possible to derive a contradiction from the inequality
$$2m < 2^r + t + 1,$$
under Case (3) and Case (4) above, considering that $2m$ is large?  (In fact, it is known that $m > {10}^{375}$.)


On OP's request, I am converting my comment into an answer.


*

*$p^k\lt m$ is equivalent to $$m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have$$\begin{align}p^k\lt m&\iff m^2-2^rt\lt m
\\\\&\iff m^2-m-2^rt\lt 0
\\\\&\iff m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\end{align}$$


*

*$(7)$ is better than $2m\lt 2^r+t+1$ since
$$\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$
holds. To see that $(8)$ holds, note that
$$\begin{align}(2)&\iff \sqrt{1+2^{r+2}t}\lt 2^r+t
\\\\&\iff 1+2^{r+2}t\lt 2^{r+1}+2^{r+1}t+t^2
\\\\&\iff (2^r-t)^2\gt 1
\\\\&\iff |2^r-t|\gt 1\end{align}$$
which does hold.



*

*We can say that $$\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$
since
$$\begin{align}(9)&\iff \bigg(\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0
\\\\&\iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\dfrac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0
\\\\&\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0
\\\\&\iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t}
\\\\&\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t)
\\\\&\iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0
\\\\&\iff (2^r-t)^2\gt 1
\\\\&\iff |2^r-t|\gt 1\end{align}$$
which does hold.



*

*It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r)$$
