(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?
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.)