Prime no.

$1991p+1=x^2 \implies 1991p=x^2-1 \implies 11\cdot 181 \cdot p=(x-1)(x+1)$

$x+1$ and $x-1$ are odd coprimes, cause, if they are even, $p$ would be a multiple of $4$ .

$11\cdot p$ won't be $2$ units away from $181$ ,for any $p$ , therefore, $p$ should be $2$ units away from $11\cdot 181=1991$ . the only option is $p=1993$