If $n = km$ is a Descartes number with quasi-Euler prime $m$, then $m < k$. (Note:  This question is tangentially related to this later one.)
Let $$\sigma(x) = \sum_{d \mid x}{d}$$ denote the sum of divisors of $x \in \mathbb{N}$, where $\mathbb{N}$ is the set of natural numbers or positive integers.
Recall that a Descartes number is an odd number $n = km$, with $1 < k$, $1 < m$, satisfying $$\sigma(k)(m+1)=2km.$$  ($m$ is called the quasi-Euler prime of $n$.)  Note that we define $\sigma(m) := m + 1$ even when $m$ is composite (that is, we pretend that $m$ is prime).
Notice that the lone Descartes number that is known is
$$\mathscr{D} = k'm' = {{3003}^2}\cdot{22021}.$$
In particular, note that:
(1) $k$ is a square.
(2) $\sigma(k)/m = 2k - \sigma(k)$
(3) $m \equiv 1 \pmod 4$
I want to prove that it must necessarily be the case that $m < k$, for a Descartes number $n = km$.
Lemma 1 If $n = km$ is a Descartes number with quasi-Euler prime $m$, then $k \neq m$.
To this end, suppose that $m = k$.  Then we have
$$\frac{\sigma(k)}{k} = \frac{2m}{m + 1} = \frac{2k}{k + 1},$$
from which it follows that
$$\sigma(k) = \frac{2k^2}{k + 1} = \frac{2k^2 - 2}{k + 1} + \frac{2}{k + 1} = \frac{2(k - 1)(k + 1)}{k + 1} + \frac{2}{k + 1} = 2(k - 1) + \frac{2}{k + 1}.$$
Since $\sigma(k)$ and $2(k - 1)$ are integers, it follows that $2/(k+1)$ is also an integer, which means that $(k + 1) \mid 2$.  This implies that $k + 1 \leq 2$, from which wet get $m = k \leq 1$.  This last inequality contradicts the condition $1 < k$, $1 < m$.
Lemma 2 If $n = km$ is a Descartes number with quasi-Euler prime $m$ and $\gcd(m,k)=1$, then
$$\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}.$$
Suppose to the contrary that $n = km$ is a Descartes number with quasi-Euler prime $m$ and $\gcd(m,k)=1$, and that $\sigma(m)/k = \sigma(k)/m$.  Then it follows that
$$\frac{\sigma(m)}{k} = \frac{\sigma(k)}{m} = r \in \mathbb{N},$$
from which we obtain
$$\frac{\sigma(m)}{k}\cdot\frac{\sigma(k)}{m} = r^2 \in \mathbb{N},$$
contradicting
$$\sigma(k)\sigma(m) = \sigma(k)(m+1) = 2km,$$
since the last two equations imply that $r^2 = 2$.
Lemma 3 If $n = km$ is a Descartes number with quasi-Euler prime $m$ and $\gcd(m, k) = 1$, then we have
(a) $\sigma(m) \neq \sigma(k)$
(b) $\sigma(m) \neq k$
(c) $\sigma(k) \neq m$
Proof of (a):  Suppose that $\sigma(k) = \sigma(m) = m + 1 \equiv 2 \pmod 4$.  This contradicts the fact that $k$ is a square, since then $\sigma(k) \equiv 1 \pmod 2$.
Proof of (b):  Suppose that $\sigma(m) = k$.  Then the even number $m + 1 = \sigma(m)$ is equal to the odd number $k$, which is a clear contradiction.
Proof of (c):  Suppose to the contrary that $\sigma(k) = m$.  Then we obtain the estimate
$$\frac{\sigma(k)}{m} + \frac{\sigma(m)}{k} = 1 + 2 = 3,$$
which contradicts the known quantity
$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$
Lemma 4 If $n = km$ is a Descartes number with quasi-Euler prime $m$ and $\gcd(m, k)=1$, then the following biconditionals hold:
$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}$$
We consider three different cases:
Case (1):
$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} = \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$
Case (1) is equivalent to $\sigma(m) = \sigma(k)$ (which is ruled out by Lemma 3 (a)) or $k = m$ (which is ruled out by Lemma 1).
Case (2):
$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k}$$
Case (2) implies the estimate
$$\frac{\sigma(m)}{k} + \frac{\sigma(k)}{m} < \frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{9 + 1}{9} + 2 = \frac{28}{9},$$
which again contradicts the known quantity
$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002.$$
Case (3):
$$\frac{\sigma(m)}{m} + \frac{\sigma(k)}{k} < \frac{\sigma(m)}{k} + \frac{\sigma(k)}{m}$$
Case (3) is equivalent to the truth of the biconditional $m < k \iff \sigma(m) < \sigma(k)$ (by virtue of Lemma 1, Lemma 2, and Lemma 3), which in turn is equivalent to the truth of the biconditional
$$m < k \iff \sigma(m) < \sigma(k) \iff \frac{\sigma(m)}{k} < \frac{\sigma(k)}{m}.$$
By Lemma 4, we have the following possibilities:
(A) $k < \sigma(k) < m < \sigma(m)$
(B) $m < \sigma(m) < k < \sigma(k)$
Note that Case (A) implies that
$$\frac{\sigma(k)}{m} = 2k - \sigma(k) < 1$$
forcing $2k - \sigma(k) = 0$ (i.e. $k$ must be perfect).  This contradicts the fact that $k$ is a square.
Hence we necessarily have Case (B), and a proof for the following theorem:
THEOREM If $n = km$ is a Descartes number with quasi-Euler prime $m$ and $\gcd(m, k) = 1$, then $k$ is not an odd almost perfect number.
QUESTIONS
(I) Can we remove the reliance of the proof on the condition
$$\frac{\sigma(k')}{m'} + \frac{\sigma(m')}{k'} = \frac{670763}{819} \approx 819.002?$$
(II) To what extent can we relax the condition $\gcd(m, k)=1$ in the THEOREM?
 A: This is a partial answer.
(II) : I think that we can prove $m\lt k$ without assuming that $\gcd(k,m)=1$. 

(1) Your proof for Lemma 1 is correct.
(2) We can prove $\frac{\sigma(m)}{k} \neq \frac{\sigma(k)}{m}$ without assuming $\gcd(k,m)=1$. 
Supposing that $\sigma(m)/k = \sigma(k)/m$ gives 
$$\frac{m+1}{k}= \frac{2k}{m+1}\implies \bigg(\frac{m+1}{k}\bigg)^2=2\implies \frac{m+1}{k}=\sqrt 2$$
which is a contradiction since LHS is rational while RHS isn't.
(3) You can prove (a) $\sigma(m) \neq \sigma(k)$ (b) $\sigma(m) \neq k$ (c) $\sigma(k) \neq m$ without assuming $\gcd(k,m)=1$ since you haven't used $\gcd(k,m)=1$ in your proof for Lemma 3.
(4) Lemma 4 is true without assuming $\gcd(k,m)=1$. The proof for $m\lt k\iff \sigma(m)\lt\sigma(k)$ is written in the question. The proof for $m\lt k\iff\frac{\sigma(m)}{k}\lt \frac{\sigma(m)}{k}$ is written in the comments below.
(5) In (A), another way to get a contradiction : We have $0\lt 2k-\sigma(k)\lt 1$ which contradicts that $2k-\sigma(k)$ is an integer. 

In conclusion, we can prove $m\lt k$ without assuming that $\gcd(k,m)=1$. 
(Moreover, I think that you can prove $m\lt k$ without using Lemma 2.)
A: Let $k$ and $m$ be odd integers greater than $1$ such that
$$(m+1)\sigma(k)=2mk.$$
Then clearly $\sigma(k)<2k$, and hence the above is equivalent to
$$m=\frac{\sigma(k)}{2k-\sigma(k)}.$$
Now the desired result that $m<k$ is equivalent to
$$\frac{\sigma(k)}{2k-\sigma(k)}<k,$$
which in turn, by clearing denominators and rearranging, is equivalent to
$$\sigma(k)<\frac{2k^2}{k+1}=2k-2+\frac{2}{k+1}.$$
In particular, as $k>1$ this is equivalent to
$$\sigma(k)\leq 2k-2.$$
As we already saw that $\sigma(k)<2k$, this is equivalent to $\sigma(k)\neq2k-1$.
So your desired result that $m<k$ is equivalent to $k$ not being an odd almost perfect number. Whether any odd almost perfect number greater than $1$ exists is an open problem, to my knowledge.
Put differently; the (integral) solutions to
$$(m+1)\sigma(k)=2mk,$$
with $k$ and $m$ both odd that do not satisfy $m<k$, are precisely the pairs
$$(k,m)=\left(k,\sigma(k)\right),$$
where $k$ is an almost perfect number. In particular $k$ is a perfect square.
A: (This is not an answer - just some comments that are too long to fit in the appropriate section.)
Note that the inequality $m < k$ is equivalent to the assertion "$k$ is not an odd almost perfect number", where $n = km$ is a Descartes number with quasi-Euler prime $m$.
This is proved in the paper "The Non-Euler Part of a Spoof Odd Perfect Number is not Almost Perfect", which is joint work between Dris and Tejada, and published in the Indian Journal in Number Theory.
