I stumbled upon the following fact recently.
$n^2 + n + p$ is prime for all integers $n \in [0, p-2]$ if and only if $4p-1$ is a Heegner number (i.e. the ring of integers of $\mathbb{Q}(i \sqrt{4p-1})$ has class number one).
The famous case is $p = 41$ by Euler, and indeed $163$ is a Heegner number. For $4p-1 = 3, 7, 11$ we have $p = 1, 2, 3$ and so it is trivial or vacuous. For $4p-1 = 19, 43, 67, 163$, we have $p = 5, 11, 17, 41$. In this second case, it is true that in the ring of integers of $\mathbb{Q}(i \sqrt{4p-1})$ all prime numbers less than $p$ are inert (this fact is used in the proof I know that $4p-1$ is a Heegner number for these values). This seems to be a promising lead, for $q < p$ being inert means that $$ \mathbb{Z}[i\sqrt{4p-1}]/(q) = \mathbb{Z}[X]/(q, X^2 + X + p) = \mathbb{F}_q[X]/(X^2 + X + p) $$ is a field, i.e. that $X^2 + X + p$ is prime in $\mathbb{F}_q[X]$. However beyond this small connection, I don't have any indication on how one would prove this and how deep this fact is.