How many integers $n$ are there such that $n^3+8$ has at most $3$ positive divisors?

$ 
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 1
\qquad\textbf{(E)}\ \text{None of the above}
$
To determine how many integers \( n \) exist such that \( n^3 + 8 \) has at most 3 positive divisors, we need to consider the structure of numbers with at most 3 divisors. 

A number can have at most 3 positive divisors in the following cases:
1. The number is of the form \( p^2 \) where \( p \) is a prime (total number of divisors = 3).
2. The number is a prime \( p \) (total number of divisors = 2).
3. The number is 1 (total number of divisors = 1), but this is not possible in this context since \( n^3 + 8 \geq 8 \).

### Case A: \( n^3 + 8 = p^2 \)
We start with the equation:
\[
n^3 + 8 = p^2
\]
This can be factored as:
\[
(n + 2)(n^2 - 2n + 4) = p^2
\]
Since \( p \) is a prime number, we have two sub-cases to consider:

#### Sub-case 1: \( n + 2 = 1 \)
\[
n + 2 = 1 \implies n = -1
\]
Substituting \( n = -1 \) into the equation:
\[
(-1)^3 + 8 = 7 \quad \text{(not a perfect square)}
\]
Thus, this sub-case does not yield a valid solution.

#### Sub-case 2: \( n + 2 = p \)
\[
n + 2 = p \implies n = p - 2
\]
Substituting \( n = p - 2 \) into the equation:
\[
(p - 2)^3 + 8 = p^2
\]
Expanding and simplifying:
\[
(p - 2)^3 + 8 = p^2 \implies p^3 - 6p^2 + 12p - 8 + 8 = p^2 \implies p^3 - 7p^2 + 12p = 0
\]
Factoring out \( p \):
\[
p(p^2 - 7p + 12) = 0
\]
Solving the quadratic equation:
\[
p^2 - 7p + 12 = 0 \implies (p - 3)(p - 4) = 0 \implies p = 3 \text{ or } p = 4
\]
For \( p = 3 \):
\[
n = 3 - 2 = 1
\]
For \( p = 4 \):
\[
n = 4 - 2 = 2 \quad \text{(not a prime number)}
\]
Thus, the valid solution from this case is \( n = 1 \).

### Case B: \( n^3 + 8 = p \)
Here, \( n^3 + 8 \) itself must be a prime number \( p \):
\[
n^3 + 8 = p
\]
We need to find \( n \) such that \( n^3 + 8 \) is a prime number.

#### Sub-case 1: \( n = -1 \)
\[
(-1)^3 + 8 = -1 + 8 = 7 \quad \text{(prime number)}
\]
Thus, \( n = -1 \) is a valid solution.

#### Sub-case 2: \( n = -3 \)
\[
(-3)^3 + 8 = -27 + 8 = -19 \quad \text{(not a prime number)}
\]
Thus, \( n = -3 \) is not a valid solution.

Combining the results from both cases, the valid solutions are:
\[
n = 1, -1
\]

The final answer is \(\boxed{2}\)