Theorem: Perfect Number which is Sum of Equal Powers of Two Numbers

$28$ is the only perfect number which is the sum of equal powers of exactly $2$ positive integers:
:$28 = 1^3 + 3^3$

Proof:
Let $N$ be an even perfect number which is the sum of equal powers of exactly $2$ positive integers.
By Theorem of Even Perfect Numbers, write:
:$N = 2^{p - 1} \paren {2^p - 1} = a^n + b^n$
where $p, 2^p - 1$ are prime, $a, b, n \in \Z_{>0}$, $n > 1$.
{{WLOG}}, let $a \le b$.
Immediately we see that $a \ne b$:
{{AimForCont}} $a = b$.
We have:
:$2^{p - 1} \paren {2^p - 1} = 2 a^n$
In particular:
:$\paren {2^p - 1} \divides a^n$
By Prime Divides Power:
:$\paren {2^p - 1}^n \divides a^n = 2^{p - 2} \paren {2^p - 1}$
which leads to a contradiction, since $n > 1$.
Next, since:
:$2^p - 1 \equiv 3 \pmod 4$
by Integer as Sum of Two Squares, $N$ is not a sum of two squares.
This leads to $n$ being odd, and $n \ge 3$.
By Sum of Two Odd Powers:
:$\paren {a + b} \divides \paren {a^n + b^n} = N$
Suppose $\paren {2^p - 1} \divides \paren {a + b}$.
We have:
{{begin-eqn}}
{{eqn | l = N
      | r = a^n + b^n
}}
{{eqn | o = >
      | r = b^n
}}
{{eqn | n = *
      | o = \ge
      | r = 4 b^2
}}
{{eqn | o = >
      | r = \paren {a + b}^2
}}
{{eqn | o = \ge
      | r = \paren {2^p - 1}^2
      | c = Absolute Value of Integer is not less than Divisors
}}
{{eqn | o = \ge
      | r = 2^{p - 1} \paren {2^p - 1}
      | c = because $2^{p - 1} \ge 1$
}}
{{end-eqn}}
If $(*)$ holds, this is a contradiction.
Since $b > a > 0$, we have:
:$b \ge 2$
Hence for $n \ge 5$:
:$b^n \ge b^5 \ge 8 b^2 \ge 4 b^2$
so $(*)$ holds.
For $n = 3$ and $b \ge 4$:
:$b^3 \ge 4 b^2$
so $(*)$ holds.
We only need to inspect the cases:
:$n = 3$, $2 \le b \le 3$, $a^3 + b^3$ is even.
This leads to the only case:
:$1^3 + 3^3 = 28$
Now suppose that $\paren {2^p - 1} \nmid \paren {a + b}$.
Then $a + b = 2^k$ for some $k \le p - 1$.
As $0 < a < b$, we have:
:$a + b \ge 3$
Hence $a, b$ are either both odd or both even.
Suppose they are both odd.
By Sum of Two Odd Powers:
:$\ds a^n + b^n = \paren {a + b} \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$
Each $\paren {-1}^j a^{n - j - 1} b^j$ is odd, and there are $n$ of them.
Therefore $\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j$ is odd.
The only odd divisors of $N$ are $1$ and $2^p - 1$.
Since $a^n + b^n > a + b$:
:$\ds \sum_{j \mathop = 0}^{n - 1} \paren {-1}^j a^{n - j - 1} b^j = 2^p - 1$
Therefore $a + b = 2^{p - 1}$.
We have:
{{begin-eqn}}
{{eqn | l = N
      | r = a^n + b^n
}}
{{eqn | o = >
      | r = b^n
}}
{{eqn | n = *
      | o = \ge
      | r = 8 b^2
}}
{{eqn | o = >
      | r = 2 \paren {a + b}^2
}}
{{eqn | r = \paren {2^{p - 1} } \paren {2^p}
}}
{{eqn | o = >
      | r = \paren {2^{p - 1} } \paren {2^p - 1}
}}
{{eqn | r = N
}}
{{end-eqn}}
If $(*)$ holds, this is a contradiction.
Since $b > a > 0$, we have:
:$b \ge 2$
Hence for $n \ge 5$:
:$b^n \ge b^5 \ge 8 b^2$
so $(*)$ holds.
For $n = 3$ and $b \ge 8$:
:$b^3 \ge 8 b^2$
so $(*)$ holds.
We only need to inspect the cases:
:$n = 3$, $2 \le b \le 7$, both $a$ and $b$ are odd, $a + b$ is a power of $2$.
This leads to the cases:
{{begin-eqn}}
{{eqn | l = 1^3 + 3^3
      | r = 28
}}
{{eqn | l = 1^3 + 7^3
      | r = 244
}}
{{eqn | l = 3^3 + 5^3
      | r = 152
}}
{{end-eqn}}
Finally, suppose they are both even.
Let $2^m$ be the largest power of $2$ that divides both $a$ and $b$, where $m \ge 1$.
Write $s = \dfrac a {2^m}$ and $t = \dfrac b {2^m}$.
Then:
:$s + t = 2^{k - m}$
:$s^n + t^n = 2^{p - 1 - m n} \paren {2^p - 1}$
At least one of $s, t$ is odd.
If the other is even, then $2^{k - m}$ is odd.
This forces $2^{k - m} = 1 = s + t$, which contradicts $a, b > 0$.
Hence $s, t$ are both odd.
By our previous analysis:
:$s + t = 2^{p - 1 - m n}$
Hence:
{{help|I cannot prove this case. Do we know if this theorem is even true? I can find nothing about it on the Internet}}