Q:

Has $\sigma\left(\sigma_0(n)^4\right)=n$ infinitely many solutions?

In this post, for integers $n\geq 1$ we consider the sum of divisors function $\sum_{d\mid n}d$ denoted as $\sigma(n)$ and the divisor-counting function $\sum_{d\mid n}1$ as $\sigma_0(n)$. Then I wondered about the solutions of $$\sigma\left((\sigma_0(n))^4\right)=n.\tag{1}$$
Here is the Wikipedia's article Divisor function dedicated to the sum of the positive divisors of an integer $n\geq 1$, that is $\sigma(n)$ and the arithmetic function $\sigma_0(n)$ (also denoted in the literature as $\tau(n)$) that counts the number of those positive divisors.
I was inspired in [1].

Question. Can you prove or refute that there exist infinitely many solutions $n\geq 1$ of $$\sigma\left((\sigma_0(n))^4\right)=n\,?$$ Many thanks.

Computational evidence. Using a Pari/GP program I know that the first few solutions of previous equation $\sigma\left(\sigma_0(n)^4\right)=n,$ are $$1 ,31 ,121 ,511, 3751$$ and $61831.$ Thus I've no enough computational evidence showing that maybe should be infinitely many solutions of $(1)$. Finally seems that previous sequence isn't in the OEIS.
References:
[1] Problem 3565, Crux Mathematicorum, Volume 37, Number 6 (2011).

A:

Let $\sigma_0(n)=m$, and we need to solve the following equation:$$\sigma_0(\sigma(m^4))=m$$
with known solutions $m=1, 2, 3, 4, 6, 12$. We have a good estimation on the upper bound of both $\sigma_0$ and $\sigma$ functions. For $\sigma_0$, $\sigma_0(x)\le x^{ \left( \frac{1.07}{\log \log x} \right) }$ and for $\sigma$, $\sigma(x)<e^\gamma x\ln \ln x+\frac{0.6483x}{\ln \ln x}$ ($\gamma$ is Euler-Mascheroni constant). For sufficiently large $x$ (at least $e^{e^{5}}$), $\sigma(x)<x^{1.1}$ and $\sigma_0(x)\le x^{0.214}$, therefore$$\sigma_0(\sigma(x^4))<\sigma_0(x^{4.4})\le x^{0.9416}<x$$
so there is only finite number of solution.

Edit June 6
I'll retry using a better upper bound, $\sigma_0(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.935}{\log \log n} \right)}$. Indeed, if $x>e^{e^{4.4}/4.4}$, then again$$\sigma_0(\sigma(x^4))<\sigma_0(x^{4.4})\le x^{0.9979}<x$$so we need to check only for $x<e^{e^{4.4}/4.4}\approx1.1\times10^8$. For such $x$, $\ln(\ln(x^4))<4.3047$ and so $\sigma(x^4)<4.3047e^\gamma x^4+\frac{0.6483x^4}{4.3047}<7.8176x^4<1.1245\times10^{33}$. Now I hope I could get a better upper bound for $\sigma_0(\sigma(x^4))$ using the fact that $\sigma(x^4)$ is odd number, but there is no researches available and the numerical result (https://oeis.org/A053624) is also limited to like $30$ digits.
Therefore, I bruted forced to find that odd number under $1.1245\times10^{33}$ can have at most $11796480$ divisors. It again gives maximum $\sigma(x^4)$ value of about $1.4706\times10^{29}$, which gives maximum divisor number of $2654208$, which gives maximum $\sigma(x^4)$ value of $3.6857\times10^{26}$, which gives maximum divisor number of $983040$, which gives maximum $\sigma(x^4)$ value of $6.8223\times10^{24}$, which gives maximum divisor number of $497664$, which gives maximum $\sigma(x^4)$ value of $4.4271\times10^{23}$, which gives maximum divisor number of $307200$, which gives maximum $\sigma(x^4)$ value of $6.3697\times10^{22}$, which gives maximum divisor number of $221184$, which gives maximum $\sigma(x^4)$ value of $1.7009\times10^{22}$, which gives maximum divisor number of $184320$, which gives maximum $\sigma(x^4)$ value of $8.1725\times10^{21}$, which gives maximum divisor number of $153600$, which gives maximum $\sigma(x^4)$ value of $3.9265\times10^{21}$, which gives maximum divisor number of $138240$, which gives maximum $\sigma(x^4)$ value of $2.5706\times10^{21}$, which gives maximum divisor number of $122880$. That is, $m \le 122880$.
Also, $2.5706\times10^{21}<3^{45}$ so $m$ cannot have prime factor greater than $45$. Computer-checking all integers under $122880$ confirms that they are indeed the only solutions.

