I am trying to characterize the solutions of the following equation involving the sum of divisors function $\sigma(n)$ and the Euler's totient function, denoted here as $\varphi(n)$.
The equation is $$\sigma\left(\frac{x(x+1)}{2}\right)=\left(2+3\varphi\left(\frac{x+2}{3}\right)\right)\left(1+3\varphi\left(\frac{x+2}{3}\right)\right).\tag{1}$$
My exploration is summarized in next claims, that I show after the question.
Question. Can you provide me a characterization of the solutions of our equation $$\sigma\left(\frac{y(y+1)}{2}\right)=\left(2+3\varphi\left(\frac{y+2}{3}\right)\right)\left(1+3\varphi\left(\frac{y+2}{3}\right)\right)$$ for positive integers $y\geq 1$? With the purpose to get such characterization, if a full characterization is impossible or very difficult, feel free to add some remarkable proposition about the solutions of our equation $(1)$. Many thanks.
Claim 1. If $x\geq 7$ is a Mersenne prime and $\frac{x+2}{3}$ is a Wagstaff prime (see this Wikipedia) then $(1)$ holds.
Remark. I've created previous equation $(1)$ thus with the intention to combine both definitions: the definition of Mersenne and Wagstaff primes.
Claim 2. Let $x\geq 7$ is a Mersenne prime solving $(1)$, then $\frac{x+2}{3}$ is a Wagstaff prime.
Sketch of proof. One has that $$3\varphi\left(\frac{x+2}{3}\right)=\frac{-3+\sqrt{1+4x(x+1)}}{2}$$ implies $\varphi((2^p+1)/3)=\frac{1}{3}(2^p+1)-1$.$\square$
Claim 3. A) If $\frac{x+2}{3}$ is a prime number such that the equation $(1)$ holds, then $\frac{x(x+1)}{2}$ is a perfect number. B) If $x\geq 7$ is prime being a solution of $(1)$, and we assume also that $\frac{x+2}{3}$ is a prime number then $\sigma(\frac{x+1}{2})=x.$