Proof

Let $n$ be an even perfect number.
Since $n$ is even, it has the factorization $n=2^{k-1}u$ with some $k > 1$, where $u > 0$ is odd.
We can calculate its sum of divisors $F(n)$ and get $$2^ku=2n=F(n)=\frac{2^k-1}{2-1}F(u)=(2^k-1)F(u).$$
It follows for the sum of divisors $F(u)$ that $$F(u)=\frac{2^ku}{2^k-1}=\frac{2^ku-u+u}{2^k-1}=\frac{(2^k-1)u+u}{2^k-1}=u+\frac u{2^k-1}.$$
Because $\frac u{2^k-1}=F(u)-u$, this fraction is a whole number.
Therefore, since $k > 1$, the number $2^k-1$ 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 $2^k-1.$
Therefore, $n=\frac{F(n)}2=2^{k-1}(2^k-1)$ where $u=2^k-1$ is a prime number.

Let $n=\frac{F(n)}2=2^{k-1}(2^k-1)$ where $p=2^k-1$ is a prime number.
We can calculate its sum of divisors $F(n)$ and get $$\sum_{d\mid n}d=F(n)=\frac{2^k-1}{2-1}\frac{p^2-1}{p-1}=(2^k-1)(p+1)=(2^k-1)2^k=2n.$$
It follows by definition that $n$ is a perfect number. if and only if $p$ is a prime number having the form $p=2^k-1$ for an $k > 1, k\in\mathbb N.$
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Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927