Which of the following cannot be the number of positive integer divisors of the number $n^2+1$, where $n$ is an integer?

$ 
\textbf{(A)}\ 2
\qquad\textbf{(B)}\ 4
\qquad\textbf{(C)}\ 6
\qquad\textbf{(D)}\ 8
\qquad\textbf{(E)}\ \text{None of above}
$
1. **Choice (A):** We need to check if there exists an integer \( n \) such that \( n^2 + 1 \) has exactly 2 positive divisors. A number with exactly 2 positive divisors must be a prime number. Let's test \( n = 1 \):
   \[
   n^2 + 1 = 1^2 + 1 = 2
   \]
   The number 2 is a prime number and has exactly 2 positive divisors (1 and 2). Therefore, choice (A) is possible.

2. **Choice (B):** We need to check if there exists an integer \( n \) such that \( n^2 + 1 \) has exactly 4 positive divisors. A number with exactly 4 positive divisors must be of the form \( p^3 \) or \( pq \) where \( p \) and \( q \) are distinct primes. Let's test \( n = 5 \):
   \[
   n^2 + 1 = 5^2 + 1 = 26
   \]
   The number 26 has the divisors 1, 2, 13, and 26, which are 4 in total. Therefore, choice (B) is possible.

3. **Choice (C):** We need to check if there exists an integer \( n \) such that \( n^2 + 1 \) has exactly 6 positive divisors. A number with exactly 6 positive divisors must be of the form \( p^5 \) or \( p^2q \) where \( p \) and \( q \) are distinct primes. Let's test \( n = 7 \):
   \[
   n^2 + 1 = 7^2 + 1 = 50
   \]
   The number 50 has the divisors 1, 2, 5, 10, 25, and 50, which are 6 in total. Therefore, choice (C) is possible.

4. **Choice (D):** We need to check if there exists an integer \( n \) such that \( n^2 + 1 \) has exactly 8 positive divisors. A number with exactly 8 positive divisors must be of the form \( p^7 \), \( p^3q \), or \( pqr \) where \( p \), \( q \), and \( r \) are distinct primes. Let's test \( n = 23 \):
   \[
   n^2 + 1 = 23^2 + 1 = 530
   \]
   The number 530 has the divisors 1, 2, 5, 10, 53, 106, 265, and 530, which are 8 in total. Therefore, choice (D) is possible.

Since all choices (A), (B), (C), and (D) are possible, the correct answer is:

\(\boxed{\text{E}}\)