All even and perfect number has the form $2^{p-1}M_p$ for some prime $p$, being $M_p$ a prime of Mersenne.
What I did was:
Suppose that $n = 2^kb$ and $\sigma(n) = 2n$. Then, $\sigma(2^kb) = (2^{k+1} - 1)\sigma(b).$ But then, once $\sigma(2n) = 2^{k+1}b$ we get that: $$2^{k+1}b = (2^{k+1} - 1)\sigma(b).$$ Even more, $\gcd(2^{k+1},2^{k+1} - 1) = 1.$ Then, $b = (2^{k+1} - 1)c,$ for some $c \in \mathbb{N}.$
I can't follow from here. Could someone help me?
Thanks!
Any other approach is welcome.
You're almost there! If we can prove that $b$ has exactly two divisors, then we're done because we know that $c$ must be $1$ and that $b$ must be prime.
Indeed, you showed that we have $$ 2^{k+1} b = 2^{k+1} (2^{k+1} - 1) c = (2^{k+1} - 1) \sigma(b) $$ therefore $$ 2^{k+1} c = \sigma(b) = b + c + \textit{stuff} = (2^{k+1} - 1) c + c + \textit{stuff} = 2^{k+1} c + \textit{stuff} $$ It follows that $\textit{stuff} = 0$, i.e. that there are no divisors of $b$ other than $c$ and $b$ itself.