(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.
Brown (2016) eventually produced a proof for the weaker inequality $p < m$. (Pomerance informed Brown that our past proofs for the estimate $p < m$ may be (inherently) flawed. See this MO question for the details of Pomerance's rebuttal.)
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)). Note: I am already too sleepy to add the details, but Dris and San Diego's proof in NNTDM for the statement "$m^2 – p^k$ is not a square" is incomplete and flawed as originally published since our proof makes use of the estimate $p < m$. (Nonetheless, it is still possible to prove the same statement, essentially by using Acquaah and Konyagin's result in (IJNT, 2012). Will submit a corrigendum for (Dris and San Diego (2020)) to NNTDM as soon as possible.)
This implies that we may write
$$m^2 – p^k = 2^r t$$
where $r \geq 2$ and $2^r \neq t$. (I have removed the condition $\gcd(2,t)=1$ since we still have not completely ruled out the possibility that $m^2 – p^k$ may be a power of two. We do know that under the assumption that $k=1$ holds, then $m^2 – p^k$ is not a power of two when $p$ is a Fermat prime.)
It is trivial to prove that $m \neq 2^r$ and $m \neq t$, so that we consider the following cases:
$$\text{Case (1): } m > t > 2^r$$
$$\text{Case (2): } m > 2^r > t$$
$$\text{Case (3): } t > m > 2^r$$
$$\text{Case (4): } 2^r > m > t$$
$$\text{Case (5): } t > 2^r > m$$
$$\text{Case (6): } 2^r > t > m$$
Note that these six (6) cases may be summarized as:
$$\text{Case (A): } m > \max(2^r, t)$$
Cases (1) and (2) are included under this Case (A).
$$\text{Case (B): } \max(2^r, t) > m > \min(2^r, t)$$
Cases (3) and (4) are included under this Case (B).
$$\text{Case (C): } \min(2^r, t) > m$$
Cases (5) and (6) are included under this Case (C).
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.
Best Answer
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 $$p^k\lt m\implies (3)\ \text{or}\ (4)\implies |2^r-t|\not=1.\quad\blacksquare$$