Every perfect number is perfect! ... or are they?

An even perfect number can be written as $2^{p-1}(2^p-1)$ where both $p$ and $2^p-1$ are prime. If we can find a positive integer $n$ and a prime $p$ such that $2^p-1$ is prime and the perfect number $X$ satisfies $X = (n+1)(n^2-n+1) = n^3+1 \; = \; 2^{p-1}(2^p-1)$ then $n$ must be odd, so that $n+1$ is even and $n^2-n+1$ is odd. If $n=1$ then $X=2$ and no prime $p$ exists. Thus $n > 1$ , so that $n^2 - n + 1 > 0$ . Thus $n+1 = 2^{p-1}$ and $n^2-n+1 = 2^p-1$ . Eliminating $p$ from these equations yields $0 \; = \; n(n-3)$ so that $n=3$ , so that $p=3$ and $X=28$ . Thus $28$ is the only even perfect number that is one more than a cube.