Primes again

Taking mod $6$ , we see that $p^2+11\equiv p^2-1\pmod{6}$ .

Now note that except for $p=2,3$ , all primes $p\equiv 1,-1\pmod{6}$ . Thus, $p^2-1\equiv (-1\text{ or }1)^2-1\equiv 0\pmod{6}$ .

Thus, $6\mid p^2+11$ .

Now, notice that we must have that $p$ cannot have any prime factors greater than $3$ . Otherwise, $p^2+11$ has at least three distinct factors and the smallest possible number of factors is $2\times2\times2=8 > 6$ .

Thus, $p^2+11=2^2\times 3=12$ or $p^2+11=2\times 3^2=18$ .

However, checking all primes $p\ge 5$ , we see that $p^2+11 \ge 36 > 18$ so there are no solutions with $p\ge 5$ .

Now we check $p=2,3$ . If $p=2$ , then $p^2+11=4+11=15$ which has $4$ factors.

We check $p=3$ . $p^2+11=20$ which has three factors. Thus, our answer is $\boxed{3}$ .