Determine all pairs of integers $(x, y)$ that satisfy equation $(y - 2) x^2 + (y^2 - 6y + 8) x = y^2 - 5y + 62$.
1. Start with the given equation:
   \[
   (y - 2) x^2 + (y^2 - 6y + 8) x = y^2 - 5y + 62
   \]

2. Expand and rearrange the terms:
   \[
   yx^2 - 2x^2 + xy^2 - 6yx + 8x = y^2 - 5y + 62
   \]

3. Combine like terms:
   \[
   yx^2 - 2x^2 + xy^2 - 6yx + 8x - y^2 + 5y - 62 = 0
   \]

4. Group terms to form perfect squares:
   \[
   [yx^2 - 2yx + y] + [4y - y^2 - 4] + [xy^2 - 4yx + 4x] + [-2x^2 + 4x - 2] = 0
   \]

5. Simplify the grouped terms:
   \[
   y(x-1)^2 - (y-2)^2 + x(y-2)^2 - 2(x-1)^2 = 56
   \]

6. Factor the equation:
   \[
   (y-2)(x-1)(x+y-3) = 56
   \]

7. Identify all factor tuples \((a, b, c)\) such that \(a \cdot b \cdot c = 56\):
   \[
   (1, 1, 56), (1, 2, 28), (1, 4, 14), (1, 7, 8), (2, 2, 14), (2, 4, 7)
   \]

8. Consider negative integers and solve for \((x, y)\) using:
   \[
   y-2 = a, \quad x-1 = b, \quad x+y-3 = c
   \]

9. Solve for each tuple:
   - For \((1, 7, 8)\):
     \[
     y-2 = 1 \implies y = 3, \quad x-1 = 7 \implies x = 8
     \]
     \[
     y-2 = 7 \implies y = 9, \quad x-1 = 1 \implies x = 2
     \]
     \[
     y-2 = 7 \implies y = 9, \quad x-1 = -8 \implies x = -7
     \]
     \[
     y-2 = 1 \implies y = 3, \quad x-1 = -8 \implies x = -7
     \]
     \[
     y-2 = -8 \implies y = -6, \quad x-1 = 7 \implies x = 8
     \]
     \[
     y-2 = -8 \implies y = -6, \quad x-1 = 1 \implies x = 2
     \]

10. Collect all solutions:
    \[
    (x, y) = (8, 3), (2, 9), (-7, 9), (-7, 3), (2, -6), (8, -6)
    \]

The final answer is \(\boxed{(8, 3), (2, 9), (-7, 9), (-7, 3), (2, -6), (8, -6)}\)