Find all positive integer triples $(x, y, z) $ that satisfy the equation $$x^4+y^4+z^4=2x^2y^2+2y^2z^2+2z^2x^2-63.$$
1. **Rewrite the given equation:**
   \[
   x^4 + y^4 + z^4 = 2x^2y^2 + 2y^2z^2 + 2z^2x^2 - 63
   \]
   We can rearrange this equation as:
   \[
   x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2 = -63
   \]

2. **Factorize the left-hand side:**
   Notice that the left-hand side can be factored using the identity for the sum of squares:
   \[
   x^4 + y^4 + z^4 - 2x^2y^2 - 2y^2z^2 - 2z^2x^2 = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + z^2x^2)
   \]
   This can be further factored as:
   \[
   (x^2 + y^2 + z^2 - xy - yz - zx)(x^2 + y^2 + z^2 + xy + yz + zx) = -63
   \]

3. **Simplify the factorization:**
   Let:
   \[
   a = x + y + z, \quad b = -x + y + z, \quad c = x - y + z, \quad d = x + y - z
   \]
   Then the equation becomes:
   \[
   abcd = 63
   \]

4. **Find the positive integer factors of 63:**
   The prime factorization of 63 is:
   \[
   63 = 3^2 \cdot 7
   \]
   The positive factors of 63 are: \(1, 3, 7, 9, 21, 63\).

5. **Check each factor:**
   - **Case 1: \(a = 63\)**
     \[
     b = c = d = 1 \implies x + y + z = 63
     \]
     This leads to a contradiction since \(x, y, z\) must be positive integers and \(x + y + z = 3\).

   - **Case 2: \(a = 21\)**
     \[
     b = 3, c = 1, d = 1 \implies x + y + z = 21
     \]
     This leads to a contradiction since \(x, y, z\) must be positive integers and \(x + y + z = 5\).

   - **Case 3: \(a = 9\)**
     \[
     b = 7, c = 1, d = 1 \implies x + y + z = 9
     \]
     Summing these two at a time gives the solution:
     \[
     (x, y, z) = (1, 4, 4) \quad \text{and permutations}
     \]

   - **Case 4: \(a = 7\)**
     \[
     b = 9, c = 1, d = 1 \implies x + y + z = 7
     \]
     This is impossible since none of these values should be greater than \(x + y + z\).

     \[
     b = 3, c = 3, d = 1 \implies x + y + z = 7
     \]
     Summing these two at a time gives the solution:
     \[
     (x, y, z) = (2, 2, 3) \quad \text{and permutations}
     \]

   - **Case 5: \(a = 3\)**
     \[
     b = 21, c = 1, d = 1 \implies x + y + z = 3
     \]
     This is impossible since none of these values should be greater than \(x + y + z\).

     \[
     b = 1, c = 3, d = 7 \implies x + y + z = 3
     \]
     This is impossible since none of these values should be greater than \(x + y + z\).

   - **Case 6: \(a = 1\)**
     This is impossible since that would make it smaller than at least one of the other three factors.

6. **Conclusion:**
   The solutions are:
   \[
   (1, 4, 4), (4, 1, 4), (4, 4, 1), (2, 2, 3), (2, 3, 2), (3, 2, 2)
   \]

The final answer is \( \boxed{ (1, 4, 4), (4, 1, 4), (4, 4, 1), (2, 2, 3), (2, 3, 2), (3, 2, 2) } \)