For odd integer $n$ it is enough to decompose:
$$n^n + 1 = (n+1)(n^{n-1} - n^{n-2} + n^{n-3} - \dots + 1).$$
Let's assume that there exists an even number $n = 2k$ such that $K = (2k)^{2k} + 1$ is divisible by $2k + 1$. Then $2k + 1$ is also a divisor of $2kK = (2k)^{2k+1} + 2k = ((2k)^{2k+1} + 1) + (2k-1)$. Since $(2k)^{2k+1} + 1 = (2k+1)((2k)^{2k} - (2k)^{2k-1} + (2k)^{2k-2} - \dots + 1)$, then $2k-1$ must also be divisible by $2k+1$. This is impossible and therefore no even number $n$ satisfies the condition.