, 2019

There is only one such integer, namely, $n = 3$; it is readily checked that $3^3 + 1 = 28$ is perfect.

If $n$ is odd, then $n^n + 1$ is even, so it is of the form $2^{p-1}(2^p - 1)$, where $p$ and $2^p - 1$ are both prime (Euler's theorem on the structure of perfect even integers). Rule out the trivial case $n = 1$, to assume $n > 1$, and write $n^n + 1 = (n + 1)(n^{n-1} - n^{n-2} + \dots - n + 1)$. Since $n$ is odd, the first factor is even and the second is odd; and since $n > 1$, the latter is greater than $1$ (simply rewrite it in the form $1 + n(n - 1)(1 + n^2 + \dots + n^{n-3})$). It follows that $n + 1 = 2^{p-1}$, so $2^p - 1 = 2n + 1$, and $n^n + 1 = (n + 1)(2n + 1) = 2n^2 + 3n + 1$ which forces $n = 3$.

We now rule out the other parity of $n$; recall that the existence of perfect odd numbers is still an open question.