How many pairs of integers $(m,n)$ are there such that $mn+n+14=\left (m-1 \right)^2$?

$ 
\textbf{a)}\ 16
\qquad\textbf{b)}\ 12
\qquad\textbf{c)}\ 8
\qquad\textbf{d)}\ 6
\qquad\textbf{e)}\ 2
$
1. Start with the given equation:
   \[
   mn + n + 14 = (m-1)^2
   \]

2. Expand and simplify the right-hand side:
   \[
   (m-1)^2 = m^2 - 2m + 1
   \]

3. Substitute this back into the original equation:
   \[
   mn + n + 14 = m^2 - 2m + 1
   \]

4. Rearrange the equation to isolate terms involving \( n \):
   \[
   mn + n + 14 - m^2 + 2m - 1 = 0
   \]

5. Combine like terms:
   \[
   mn + n - m^2 + 2m + 13 = 0
   \]

6. Factor out \( n \) from the left-hand side:
   \[
   n(m + 1) - m^2 + 2m + 13 = 0
   \]

7. Rearrange to use Simon's Favorite Factoring Trick:
   \[
   n(m + 1) = m^2 - 2m - 13
   \]

8. Add and subtract 3 on the right-hand side to complete the factorization:
   \[
   n(m + 1) = m^2 - 2m + 3 - 16
   \]

9. Factor the right-hand side:
   \[
   n(m + 1) = (m - 1)(m + 3) - 16
   \]

10. Rearrange to isolate the product of two factors:
    \[
    n(m + 1) + (m - 1)(m + 3) = -10
    \]

11. Factor the left-hand side:
    \[
    (m + 1)(n - m + 3) = -10
    \]

12. Identify the pairs of factors of \(-10\):
    \[
    (-10, 1), (1, -10), (-1, 10), (10, -1), (2, -5), (-5, 2), (-2, 5), (5, -2)
    \]

13. For each pair \((a, b)\), solve for \(m\) and \(n\):
    \[
    m + 1 = a \quad \text{and} \quad n - m + 3 = b
    \]

    This gives:
    \[
    m = a - 1 \quad \text{and} \quad n = b + m - 3
    \]

    Substituting \(m = a - 1\) into \(n = b + m - 3\):
    \[
    n = b + (a - 1) - 3 = b + a - 4
    \]

14. Verify each pair:
    - For \((-10, 1)\):
      \[
      m = -11, \quad n = 1 - 10 - 4 = -13
      \]
    - For \((1, -10)\):
      \[
      m = 0, \quad n = -10 + 1 - 4 = -13
      \]
    - For \((-1, 10)\):
      \[
      m = -2, \quad n = 10 - 1 - 4 = 5
      \]
    - For \((10, -1)\):
      \[
      m = 9, \quad n = -1 + 10 - 4 = 5
      \]
    - For \((2, -5)\):
      \[
      m = 1, \quad n = -5 + 2 - 4 = -7
      \]
    - For \((-5, 2)\):
      \[
      m = -6, \quad n = 2 - 5 - 4 = -7
      \]
    - For \((-2, 5)\):
      \[
      m = -3, \quad n = 5 - 2 - 4 = -1
      \]
    - For \((5, -2)\):
      \[
      m = 4, \quad n = -2 + 5 - 4 = -1
      \]

15. Each pair \((a, b)\) gives a unique pair \((m, n)\), resulting in 8 pairs.

The final answer is \(\boxed{8}\)