How many primes $p$ are there such that the number of positive divisors of $p^2+23$ is equal to $14$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 3
\qquad\textbf{(E)}\ \text{None of above}
$
To solve the problem, we need to find the number of prime numbers \( p \) such that the number of positive divisors of \( p^2 + 23 \) is equal to 14. 

1. **Understanding the divisor function**:
   The number of positive divisors of a number \( n \) can be determined from its prime factorization. If \( n \) has the prime factorization \( n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k} \), then the number of positive divisors of \( n \) is given by:
   \[
   d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   We are given that \( d(p^2 + 23) = 14 \). The number 14 can be factored as \( 14 = 1 \times 14 \) or \( 14 = 2 \times 7 \).

2. **Case 1: \( p^2 + 23 \) is a 13th power of a prime**:
   If \( p^2 + 23 = q^{13} \) for some prime \( q \), then \( p^2 + 23 \) must be a 13th power of a prime. However, since \( p^2 + 23 \) is even (as \( p \) is a prime and \( p^2 \) is odd for all odd primes), \( q \) must be 2. But \( 2^{13} = 8192 \), and \( p^2 + 23 = 8192 \) implies \( p^2 = 8169 \), which is not a perfect square. Therefore, this case is not possible.

3. **Case 2: \( p^2 + 23 \) has the form \( p_1^6 p_2 \)**:
   If \( p^2 + 23 = p_1^6 p_2 \) for some primes \( p_1 \) and \( p_2 \), then the number of divisors is \( (6+1)(1+1) = 14 \). Since \( p^2 + 23 \) is even, one of the primes must be 2. Let's assume \( p_1 = 2 \):
   \[
   p^2 + 23 = 2^6 p_2 = 64 p_2
   \]
   This implies:
   \[
   p^2 + 23 = 64 p_2
   \]
   Rearranging, we get:
   \[
   p^2 = 64 p_2 - 23
   \]
   For \( p^2 \) to be a perfect square, \( 64 p_2 - 23 \) must be a perfect square. Let's check for small prime values of \( p_2 \):

   - If \( p_2 = 3 \):
     \[
     p^2 = 64 \cdot 3 - 23 = 192 - 23 = 169 \implies p = 13
     \]
     Since 13 is a prime, this is a valid solution.

   - If \( p_2 = 5 \):
     \[
     p^2 = 64 \cdot 5 - 23 = 320 - 23 = 297
     \]
     297 is not a perfect square.

   - If \( p_2 = 7 \):
     \[
     p^2 = 64 \cdot 7 - 23 = 448 - 23 = 425
     \]
     425 is not a perfect square.

   - If \( p_2 = 11 \):
     \[
     p^2 = 64 \cdot 11 - 23 = 704 - 23 = 681
     \]
     681 is not a perfect square.

   - If \( p_2 = 17 \):
     \[
     p^2 = 64 \cdot 17 - 23 = 1088 - 23 = 1065
     \]
     1065 is not a perfect square.

   Therefore, the only valid solution is when \( p = 13 \).

Conclusion:
There is only one prime \( p \) such that the number of positive divisors of \( p^2 + 23 \) is equal to 14.

The final answer is \(\boxed{1}\)