Eleven primes

If $p\not=2$ , $p\equiv \pm 1\mod 4$ and $p^2\equiv 1\mod 4$ . Si $p\not=3 \Rightarrow p\equiv \pm 1\mod 3$ and $p^2\equiv 1\mod 3$ . Therefore, if $p\geq5$ , then $3,4$ and $12$ divides $p^2+11$ .

$12$ has six positive divisors so $p^2+11$ has at least one more divisor.

$p=2 \Rightarrow 2^2+11=15$ has four divisors.

$p=3 \Rightarrow 3^2+11=20$ has six divisors divisors.

$\boxed{p=3}$

just try p from the lowest possible prime numbers

if $p = 2$ then $p^2 + 11 = 15$ which only have 1,3,5, and 15 as its positive divisors (4 divisors) .

else if $p = 3$ then $p^2 + 11 = 20$ which have 1,2,4,5,10,and 20 as its positive divisors (6 divisors)