From the equation $\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1)$ involving arithmetic functions to a characterization of Mersenne exponents In this post we denote the Euler's totient function that counts the number of positive integers $1\leq k\leq n$ such that $\gcd(k,n)=1$ as $\varphi(n)$, and the sum of divisors function $\sum_{1\leq d\mid n}d$ as $\sigma(n)$. As reference I add the Wikipedia Mersenne prime) that refers the well-known definition that a prime $p$ is a Mersenne exponent if $2^p-1$ is prime. This prime constellation corresponds to the entry A000043 from the On-Line Encyclopedia of Integer Sequences.
From previous definition and the calculations of previous arithmetic function $\varphi(n)$ and $\sigma(n)$ it is easy to check the proof of the following claim.
Claim. If $x=p$ is (a prime) such that $y=2^p-1$ is (also) prime, then $(x,y)$ solves the equation $$\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1).\tag{1}$$
We propose the following conjecture inspired in previous claim, from the substitution $$y=2\cdot 2^{\varphi(x)}-1=2^{1+\varphi(x)}-1.\tag{2}$$
Conjecture. Let $x\geq 1$ be an integer that satisfies
$$\sigma(x^{\sigma(2^{1+\varphi(x)}-1)-1})=\frac{1}{\varphi(x)}(x^{2^{1+\varphi(x)}}-1),\tag{3}$$
then $x$ is a Mersenne exponent.

Question. What work can be done with the purpose to prove or refute previous Conjecture? Many thanks.

Computational evidence.  You can check in the web Sage Cell Server this line written in Pari/GP
for(x=1,16,if(sigma(x^(sigma(2*2^(eulerphi(x))-1 )-1))==(x^(2*2^(eulerphi(x))-1 +1)-1)/eulerphi(x) ,print(x)))
just copy and paste it to evaluate in the web choosing as Language the option GP.
 A: The conjecture is true.
Proof : 
$x=1$ does not satisfy $(3)$, and $x=2$ satisfies $(3)$. In the following, $x\ge 3$.
As you already noticed, $x$ has to be a square-free integer.
Then, letting $x=\displaystyle\prod_{k=1}^{n}p_k$ with $n=\omega(x)$, we get
$$\prod_{k=1}^{n}\bigg({p_k}^{\sigma(2^{1+\varphi(x)}-1)}-1\bigg)=-1+\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$
Suppose here that $2^{1+\varphi(x)}-1$ is a composite number. 
Here, using the following facts :


*

*If $N$ is a composite number, then $\sigma(N)\ge 1+\sqrt N+N$.

*If $N\ge 3$, then $\varphi(N)\ge 2$.

*If $m\ge 2$ and $y\gt 0$, then $m^{2+y}-1\ge m^{1+y}$.
we get
$$\begin{align}-1&=\prod_{k=1}^{n}\bigg({p_k}^{\sigma(2^{1+\varphi(x)}-1)}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}
\\\\&\ge \prod_{k=1}^{n}\bigg({p_k}^{1+\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}-1}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}
\\\\&=\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+\varphi(x)}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}
\\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{\sqrt{2^{1+2}-1}+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}
\\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{2+2^{1+\varphi(x)}}-1\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}
\\\\&\ge\prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}\end{align}$$
from which we have
$$-1\ge \prod_{k=1}^{n}\bigg({p_k}^{1+2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$
which is impossible since the RHS is positive.
So, we see that $2^{1+\varphi(x)}-1$ is a prime number.
Then, we get
$$\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg)=-1+\prod_{k=1}^{n}{p_k}^{2^{1+\varphi(x)}}$$
Suppose here that $n\ge 2$. Then, we get
$$\begin{align}1&=\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\prod_{k=1}^{n}\bigg({p_k}^{2^{1+\varphi(x)}}-1\bigg)
\\\\&\ge {p_n}^{2^{1+\varphi(x)}}\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)-\bigg({p_n}^{2^{1+\varphi(x)}}-1\bigg)\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)
\\\\&=\prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)\end{align}$$
from which we have
$$1\ge \prod_{k=1}^{n-1}\bigg({p_k}^{2^{1+\varphi(x)}}\bigg)$$
which is impossible since the RHS is larger than $1$.
So, we have to have $\omega(x)=n=1$, and $x$ has to be a prime number.
Therefore, $x$ has to be a Mersenne exponent. $\quad\blacksquare$
