Necessity: Let $p=2^k-1$ be prime and $n=2^{k-1}p$ We have that \begin{eqnarray*} \sigma(n) & = & \sigma(2^{k-1}p)\\ & = & \sigma(2^{k-1}) \sigma(p)\\ & = & (2^k-1)(p+1)\\ & = & (2^k-1)2^k\\ & = & 2n, \end{eqnarray*}which shows that $n$ is perfect.

Sufficiency: Assume $n$ is an even perfect number. Write $n=2^{k-1}m$ for some odd $m$ and some $k>1$ Then we have $\gcd(2^{k-1},m)=1$ Thus, $$ \sigma(n)=\sigma(2^{k-1}m)=\sigma(2^{k-1})\sigma(m)=(2^k-1)\sigma(m). $$ Since $n$ is perfect, $\sigma(n)=2n$ by definition. Therefore, $\sigma(n)=2n=2^km$ Piecing together the two formulas for $\sigma(n)$ yields $$ 2^km=(2^k-1)\sigma(m). $$ Thus, $(2^k-1)\mid 2^km$ which forces $(2^k-1)\mid m$ Write $m=(2^k-1)M$ Note that $1\le M<m$ From above, we have: \begin{eqnarray*} 2^km & = & (2^k-1)\sigma(m) \\ 2^k(2^k-1)M & = & (2^k-1)\sigma(m) \\ 2^kM & = & \sigma(m) \end{eqnarray*}Since $m\mid m$ by definition of divides and $M\mid m$ by assumption, we have $$ 2^kM=\sigma(m)\geq m+M=2^kM, $$ which forces $\sigma(m)=m+M$ Therefore, $m$ has only two positive divisors, $m$ and $M$ Hence, $m$ must be prime, $M=1$ and $m=(2^k-1)M=2^k-1$ from which the result follows.