Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions?

Question

This question was posted in MSE in early August 2020. It did garner several upvotes, but did not receive any responses. I have therefore cross-posted it here, hoping that it gets answered.

Let $\sigma(x)$ denote the classical sum of divisors of the positive integer $x$.

Here is my question:

Does the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ have any odd solutions?

MY ATTEMPT

I tried searching for solutions to the equation in Sage Cell Server, in the range $1 < x \leq {10}^6$, here is the Pari-GP code:

for(x=1, 1000000, if(sigma(sigma(x^2))==2*x*sigma(x),print(x,factor(x))))

Here are the results:

9516[2, 2; 3, 1; 13, 1; 61, 1]
380640[2, 5; 3, 1; 5, 1; 13, 1; 61, 1]

Note that both results obtained $x_1 = 9516$ and $x_2 = 380640$ are even.

The Pari-GP interpreter of Sage Cell Server crashes as soon as a search limit of ${10}^7$ is specified.

CONJECTURE

The equation $\sigma(\sigma(x^2)) = 2x\sigma(x)$ does not have any odd solutions.

Alas, I have no proof.

SOME PARTIAL RESULTS

Per Will Jagy's answer (and a subsequent comment by Erick Wong) to the following MSE question, we have the conjectured (sharp?) upper bound $$\frac{\sigma(x^2)}{x\sigma(x)} \leq \prod_{\rho}{\frac{{\rho}^2 + \rho + 1}{{\rho}^2 + \rho}} = \frac{\zeta(2)}{\zeta(3)}.$$

Claim: $\sigma(x^2) \neq p^r$ if prime $p \geq 5$.

Suppose to the contrary that the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ has an odd solution, and that $\sigma(x^2) = p^r$ for some prime $p \geq 5$.

Then we have $$\frac{\sigma(p^r)}{p^r}=\frac{\sigma(\sigma(x^2))}{\sigma(x^2)}=\frac{2x\sigma(x)}{\sigma(x^2)} \geq \frac{2\zeta(3)}{\zeta(2)} \approx 1.4615259388.$$ But we know that the abundancy index $\sigma(p^r)/p^r$ is bounded above by $$\frac{\sigma(p^r)}{p^r} < \frac{p}{p - 1} \leq \frac{5}{4}.$$ This is a contradiction.

Claim: $\sigma(x^2) \neq p^r q^s$ if primes $p \geq 5$ and $q \geq 7$.

Suppose to the contrary that the equation $\sigma(\sigma(x^2))=2x\sigma(x)$ has an odd solution, and that $\sigma(x^2) = p^r q^s$ for some primes $p \geq 5$ and $q \geq 7$.

Then we have $$\frac{\sigma(p^r)}{p^r}\cdot\frac{\sigma(q^s)}{q^s}=\frac{\sigma(\sigma(x^2))}{\sigma(x^2)}=\frac{2x\sigma(x)}{\sigma(x^2)} \geq \frac{2\zeta(3)}{\zeta(2)} \approx 1.4615259388.$$ But we know that the product of abundancy indices $(\sigma(p^r)/p^r)(\sigma(q^s)/q^s)$ is bounded above by $$\frac{\sigma(p^r)}{p^r}\cdot\frac{\sigma(q^s)}{q^s} < \frac{p}{p - 1}\cdot\frac{q}{q - 1} \leq \frac{5}{4}\cdot\frac{7}{6} = \frac{35}{24} = 1.458\overline{333}.$$ This is a contradiction.

Edit (August 27, 2020 - 4:21 PM Manila time)

MOTIVATION FOR THE PROBLEM

Slowak (1999) proved that an odd perfect number $N = u^t v^2$ (with special prime $u$ satisfying $u \equiv t \equiv 1 \pmod 4$ and $\gcd(u,v)=1$) must be of the form $$\frac{u^t \sigma(u^t)}{2}\cdot{d},$$ where $d > 1$.

Dris (2017) showed further that $d$ must have the form $$\frac{D(v^2)}{\sigma(u^{t-1})}=\gcd(v^2,\sigma(v^2))=\frac{\sigma(v^2)}{u^t}=\frac{v^2}{\sigma(u^t)/2},$$ where $D(y)=2y-\sigma(y)$ is the deficiency of $y$.

If the odd perfect number $N = u^t v^2$ is of the form $$\frac{u^t \sigma(u^t)}{2}\cdot{v},$$ then by the results of Slowak and Dris, we have the equations $$\sigma(v^2) = u^t v$$ and $$2v = \sigma(u^t),$$ which led me to my question in this post.