Solution:

For

 k = 1, n = 7 n2 - k = 48 = 3 . 24,

 k = 2, n = 235 n2 - k = 55227 = 23 . 74,

 k = 3, n = 4936 n2 - k = 24364093 = 17 . 374.

Therefore, k > 4. Let us prove that the number of divisors is not 10. If n2 - 4 has exactly 10 divisors, then n2 - 4 = p9 or n2 - 4 = p . q4 for some prime numbers p and q. In the first case, (n - 2)(n + 2) = p9; if n = 3 contradiction, if n- 2 > 1, then p divides both n- 2 and n + 2, and hence p divides n + 2 - (n - 2) = 4. But the only possibility p = 2 fails. In the second case, (n- 2)(n + 2) = pq4, as in the first case, if q divides both n- 2 and n + 2, then q = 2, therefore n = 2m. By substitution n = 2m we get m2 - 1 = (m - 1)(m + 1) = 4p, hence m is odd. But if m is odd, then m2 - 1 = 0 (mod 8) and 4p = 0 (mod 8), and p is even, contradiction. Therefore, q must divide only one of two numbers n - 2 and n + 2, words, one of this number is q4, other one p. We have two cases:

1. n + 2 = p, n - 2 = q4 and p = 4 + q4. q is not 5 (4 + 54 = 629 is not prime). Therefore, q4 = 1 (mod 5) and p = 4 + q4 = 0 (mod 5), and p = 5. Contradiction.

2. n + 2 = q4, n- 2 = p and p = q4 - 4 = (q2 - 2)(q2 + 2). Since q > 1, p is not prime. Contradiction.