Consider the following inequality:
$$\frac{6}{{\pi}^2} < \frac{\sigma(n)\phi(n)}{n^2} \lt 1.$$
For the even perfect numbers, this inequality seems to converge to 1. Here's the following ($5$ decimal) values for the first $5$ perfect numbers:
$$0.66666,0.85714,0.96774,0.99212,0.99987$$
Question: Does the convergence to $1$, if it can be proven, show that there are an infinite number of even perfect numbers?
On
(Note: This is too long for a comment and complements player3236's answer.)
Consider a hypothetical odd perfect number $N$.
From the following source: Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds $$\dfrac{240}{217\zeta(3)} < \dfrac{\sigma(N)\varphi(N)}{N^2} < 1,$$ where we have the rational approximation $$\dfrac{240}{217\zeta(3)} \approx 0.92008188672520641697672952496390495972393334311564395865.$$
Note that we also have $$\dfrac{6}{{\pi}^2} \approx 0.607927101854026628663276779258365833426152648.$$
Consider the even perfect numbers. (We don't know if odd ones exist)
By Euclid-Euler Theorem, $n = 2^{p-1} (2^p - 1)$ for some prime $p$ and $2^p-1$.
$$\phi(n)=\phi(2^{p-1})\phi(2^p-1) = 2^{p-2}(2^p-2)$$
$$\frac {\sigma (n) \phi(n)}{n^2} = \frac {(2n)(2^{p-2}(2^p-2))}{n (2^{p-1} (2^p - 1))} = 1-\frac 1{2^p-1}$$
hence this value does converge to $1$ (for the even perfect numbers).
However, as the existence of an infinite number of perfect numbers (or Mersenne primes) is not proven yet, I would presume this argument leads nowhere.