Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms? | A proposed proof (?) for the nonexistence of odd perfect numbers

(Note 1: The following blog post is taken verbatim from this MSE question.)

(Note 2: This was cross-posted from MO, because it was not well-received there.  Will delete the MO post in a few.)

Observe that an even perfect number $M = (2^p - 1)\cdot{2^{p - 1}}$ and an odd perfect number $N = q^k n^2$ have similar multiplicative forms.  (Indeed, it is conjectured that $k=1$, and this prediction goes back to Descartes ($1638$).)

Here is my initial question:

INITIAL QUESTION
Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms?

I ask because of a related query here.

Indeed, if we could show that $$\left|2^r - t\right| = 1,$$
for $r$ and $t$ satisfying $n^2 - q^k = 2^r t$ and $\gcd(2, t)=1$, then we would have a proof for $n < q^k$, which together with Brown's estimate $q < n$ ($2016$) would yield a refutation of Descartes's Conjecture that $k = 1$.

Note that, for even perfect numbers, we actually have
$$(2^p - 1) - 2^{p-1} = 2^{p-1} - 1 = \bigg(2^{\frac{p-1}{2}} + 1\bigg)\bigg(2^{\frac{p-1}{2}} - 1\bigg) = ab,$$
where the factorization works whenever $M \neq 6$.  (Note that $\gcd(a, b) = 1$.)

We compute that
$$\left|a - b\right| = 2.$$

DISCUSSION

In the hyperlinked MO question, the following (summarized) cases were considered for odd perfect numbers, which we now examine for even perfect numbers:

Case 1:
$$2^{\frac{p-1}{2}} < \min(a,b) = 2^{\frac{p-1}{2}} - 1$$
Notice that Case 1 clearly does not hold.

Case 2:
$$\min(a,b) = 2^{\frac{p-1}{2}} - 1 < 2^{\frac{p-1}{2}} < \max(a,b) = 2^{\frac{p-1}{2}} + 1$$
Notice that Case 2 clearly holds.

It follows (from mimicking the resulting inequality $q^k < n\cdot{\left|2^r - t\right|}$ for odd perfect numbers) that
$$2^p - 1 < {2^{\frac{p-1}{2}}}\cdot{\left|a - b\right|} = {2^{\frac{p-1}{2}}}\cdot{2} = {2^{\frac{p+1}{2}}}.$$

This last inequality implies that
$$p < -\frac{2\bigg(\log(2) - \log(\sqrt{2} + \sqrt{6})\bigg)}{\log(2)} \approx 1.89997,$$
which is a contradiction to $p \geq 2$.

Case 3:
$$\max(a,b) = 2^{\frac{p-1}{2}} + 1 < 2^{\frac{p-1}{2}}$$
Notice that Case 3 clearly does not hold.
 
Note that for even perfect numbers, we do have
$$2^{\frac{p-1}{2}} < 2^p - 1$$
which "mimics" the conjecture $n < q^k$ for odd perfect numbers above.

Here is my final question for this post:

FINAL QUESTION
Does the exhaustion of all possible cases in the DISCUSSION section essentially disprove the existence of odd perfect numbers?