, 2019

Solution:
$n^{3}+n^{2}+n+1 = (n+1)(n^{2}+1)$. Note that $\gcd(n^{2}+1, n+1) = \gcd(2, n+1) = 1$ or $2$, and since $n^{2}+1$ is not a perfect square for $n \geq 1$, we must have $n^{2}+1 = 2p^{2}$ and $n+1 = 2q^{2}$ for some integers $p$ and $q$. The first equation is a variant of Pell's equation, which (either by brute-forcing small cases or using the known recurrence) gives solutions $(n, p) = (1, 1), (7, 5), \ldots$ Incidentally, both smallest solutions $n=1$ and $n=7$ allow an integer solution to the second equation, so $n=7$ is the second smallest integer that satisfies the condition.