Solutions of $\varphi\left(n^n\sigma(n)\right)=\varphi(n^{n+1})$, where $\sigma(n)$ is the sum of divisors and $\varphi(n)$ the Euler's totient I would like to know a full characterization of the solutions of next equation involving the sum of divisors function $\sigma(m)=\sum_{d\mid m}d$ and the Euler's totient function denoted as $\varphi(m)$
$$\varphi\left(n^n\sigma(n)\right)=\varphi(n^{n+1}).\tag{1}$$
The equation arises from my experiments with odd perfect numbers, since one has the following claim.
Claim. If there exists an odd perfect number then satisfies $(1)$ (and as a consequence we can get also an statement invoking Euler-Fermat $2^{\varphi(n^n\sigma(n))}\equiv 1\text{ mod }n^{n+1}$, I know the first few solutions of this congruence).
Just a comment is that I was playing with more reasonable powers before this $n^n$. 
Proof. The proof is easy from the fact that the Euler's totient function is multiplicative, $\varphi\left(n^n\sigma(n)\right)=\varphi(2n^{n+1})=\varphi(n^{n+1}).$$\square$
Now, it was fun when I ran a program to know what integers satisfy our equation $(1)$. The sequence starts as $1, 2, 8, 128$, that are the terms that I can compute with my program. But due my belief I suspect that if fact every term from the OEIS A058891 satisfy $(1)$.

Question. I would like to know a full characterization (I am saying what reasonable work can be done) of solutions of previous equation $(1)$. I know the Claim and I suspect that the integers of the form $$2^{2^{\lambda-1}-1}\tag{2}$$
  satisfy our equation, where $\lambda$ runs over integers $\geq 1$. What can you prove about it? Can you prove that $(2)$ satisfies our equation (I know that it should be easy but is required patience)? Are there more solutions? Many thanks.

Please if you know the equation $(1)$ from the literature please answer this as a reference request and I try to find and read those facts from the literature.
 A: About $2^{2^{\lambda-1}-1}$:
Assume $2^a$ satisfies the equation. Then, since $\sigma(2^a)=2^{a+1}-1$, we have that
$$\varphi\left(2^{a2^a}\left(2^{a+1}-1\right)\right)=\varphi\left(2^{a2^a+a}\right).$$
Since $\varphi(mn)=\varphi(m)\varphi(n)$ when $\gcd(m,n)=1$ and $\phi(2^a)=2^{a-1},$ we have that then
$$2^{a2^a-1}\varphi\left(2^{a+1}-1\right) = 2^{a2^a+a-1}.$$
$$\varphi\left(2^{a+1}-1\right) = 2^a.$$
You wish to show that 
$$\varphi\left(2^{2^k}-1\right) = 2^{2^k-1}.$$
Let
$$F_n=2^{2^n}+1.$$
Then
$$2^{2^n}-1=\left(2^{2^{n-1}}-1\right)\left(2^{2^{n-1}}-1\right),$$
so we have inductively that
$$2^{2^n}-1 = F_0\cdots F_{n-1}.$$
Since all Fermat numbers are coprime, this holds iff we have that
$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) = \prod_{k=0}^{n-1} \left(2^{2^k}\right).$$
Indeed, since $\varphi(x)\leq x-1$, with equality holding iff $x$ is prime, we have that 
$$\prod_{k=0}^{n-1} \varphi\left(2^{2^k}+1\right) \leq \prod_{k=0}^{n-1} \left(2^{2^k}\right),$$
with equality holding iff each of the first $n-1$ Fermat numbers are prime. Thus, your conjecture is equivalent to that made by Fermat that they are all prime, and thus fails first at $n=6$, and similarly everywhere thereafter (as $F_5$ is not prime). 
