For how many primes $p$, $|p^4-86|$ is also prime?

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
1. **Check for \( p = 5 \):**
   \[
   |p^4 - 86| = |5^4 - 86| = |625 - 86| = 539
   \]
   Since \( 539 = 7^2 \times 11 \), it is not a prime number.

2. **Assume \( p \neq 5 \):**
   - By Fermat's Little Theorem, for any prime \( p \neq 5 \), we have:
     \[
     p^4 \equiv 1 \pmod{5}
     \]
   - This implies:
     \[
     p^4 - 86 \equiv 1 - 86 \equiv -85 \equiv 0 \pmod{5}
     \]
     Therefore, \( 5 \) divides \( |p^4 - 86| \).

3. **Since \( |p^4 - 86| \) is a prime number and divisible by 5, it must be equal to 5:**
   - This gives us two cases:
     \[
     p^4 - 86 = 5 \quad \text{or} \quad 86 - p^4 = 5
     \]

4. **Solve \( p^4 - 86 = 5 \):**
   \[
   p^4 = 91
   \]
   Since 91 is not a perfect fourth power of any integer, there is no solution in this case.

5. **Solve \( 86 - p^4 = 5 \):**
   \[
   86 - 5 = p^4 \implies p^4 = 81
   \]
   \[
   p = \sqrt[4]{81} = 3
   \]
   Since \( 3 \) is a prime number, this is a valid solution.

6. **Verify the solution:**
   \[
   |3^4 - 86| = |81 - 86| = 5
   \]
   Since 5 is a prime number, \( p = 3 \) is indeed a valid solution.

Conclusion:
There is only one prime \( p \) such that \( |p^4 - 86| \) is also a prime.

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