Determine all pairs of positive integers $(x, y)$ such that:
$$x^4 - 6x^2 + 1 = 7\cdot 2^y$$
To determine all pairs of positive integers \((x, y)\) such that:
\[ x^4 - 6x^2 + 1 = 7 \cdot 2^y \]

1. **Rewrite the equation:**
   \[ x^4 - 6x^2 + 1 = 7 \cdot 2^y \]
   Let \( z = x^2 \). Then the equation becomes:
   \[ z^2 - 6z + 1 = 7 \cdot 2^y \]

2. **Complete the square:**
   \[ z^2 - 6z + 1 = (z - 3)^2 - 9 + 1 = (z - 3)^2 - 8 \]
   So, the equation can be rewritten as:
   \[ (z - 3)^2 - 8 = 7 \cdot 2^y \]
   \[ (z - 3)^2 = 7 \cdot 2^y + 8 \]

3. **Analyze the parity:**
   For \( y > 3 \), the right-hand side (RHS) \( 7 \cdot 2^y + 8 \) is even, but not a perfect square because \( 7 \cdot 2^y \) is divisible by \( 2^3 \) and adding 8 makes it not a perfect square. Therefore, \( y \) must be less than or equal to 3.

4. **Test small values of \( y \):**
   - For \( y = 1 \):
     \[ (z - 3)^2 = 7 \cdot 2^1 + 8 = 14 + 8 = 22 \]
     \[ (z - 3)^2 = 22 \]
     This is not a perfect square.

   - For \( y = 2 \):
     \[ (z - 3)^2 = 7 \cdot 2^2 + 8 = 28 + 8 = 36 \]
     \[ (z - 3)^2 = 36 \]
     \[ z - 3 = \pm 6 \]
     \[ z = 9 \text{ or } z = -3 \]
     Since \( z = x^2 \) and \( z \) must be positive, \( z = 9 \).
     \[ x^2 = 9 \]
     \[ x = 3 \]

   - For \( y = 3 \):
     \[ (z - 3)^2 = 7 \cdot 2^3 + 8 = 56 + 8 = 64 \]
     \[ (z - 3)^2 = 64 \]
     \[ z - 3 = \pm 8 \]
     \[ z = 11 \text{ or } z = -5 \]
     Since \( z = x^2 \) and \( z \) must be positive, \( z = 11 \).
     \[ x^2 = 11 \]
     This does not yield an integer solution for \( x \).

Therefore, the only solution is when \( y = 2 \) and \( x = 3 \).

The final answer is \( \boxed{ (x, y) = (3, 2) } \)