Proof
(related to Proposition: Even Perfect Numbers)
"$\Rightarrow$"
 Let $n$ be an even perfect number.
 Since $n$ is even, it has the factorization $n=2^{k1}u$ with some $k > 1$, where $u > 0$ is odd.
 We can calculate its sum of divisors $F(n)$ and get
$$2^ku=2n=F(n)=\frac{2^k1}{21}F(u)=(2^k1)F(u).$$
 It follows for the sum of divisors $F(u)$ that
$$F(u)=\frac{2^ku}{2^k1}=\frac{2^kuu+u}{2^k1}=\frac{(2^k1)u+u}{2^k1}=u+\frac u{2^k1}.$$
 Because $\frac u{2^k1}=F(u)u$, this fraction is a whole number.
 Therefore, since $k > 1$, the number $2^k1$ is a divisor of $u$, and it is even.
 The sum of divisors $F(u)$ is therefore the sum of $u$ and some proper divisor(s).
 But $u$ has no even divisors, since it is odd. Therefore $u$ is a prime number of the form $2^k1.$
 Therefore, $n=\frac{F(n)}2=2^{k1}(2^k1)$ where $u=2^k1$ is a prime number.
"$\Leftarrow$"
 Let $n=\frac{F(n)}2=2^{k1}(2^k1)$ where $p=2^k1$ is a prime number.
 We can calculate its sum of divisors $F(n)$ and get
$$\sum_{d\mid n}d=F(n)=\frac{2^k1}{21}\frac{p^21}{p1}=(2^k1)(p+1)=(2^k1)2^k=2n.$$
 It follows by definition that $n$ is a perfect number.
if and only if $p$ is a prime number having the form $p=2^k1$ for an $k > 1, k\in\mathbb N.$
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927