Find all positive integer solutions $(x,y,z,n)$ of equation $x^{2n+1}-y^{2n+1}=xyz+2^{2n+1}$, where $n\ge 2$ and $z \le 5\times 2^{2n}$.
1. Given the equation \( x^{2n+1} - y^{2n+1} = xyz + 2^{2n+1} \), we need to find all positive integer solutions \((x, y, z, n)\) where \( n \geq 2 \) and \( z \leq 5 \times 2^{2n} \).

2. First, observe that \( x - y \) must be even because the left-hand side \( x^{2n+1} - y^{2n+1} \) is odd (since \(2n+1\) is odd) and the right-hand side \( xyz + 2^{2n+1} \) is also odd (since \(2^{2n+1}\) is even and \(xyz\) is odd). Therefore, \( x - y \geq 2 \).

3. Consider the factorization of the left-hand side:
   \[
   x^{2n+1} - y^{2n+1} = (x - y)(x^{2n} + x^{2n-1}y + \cdots + y^{2n})
   \]
   Since \( x - y \geq 2 \), we have:
   \[
   x^{2n+1} - y^{2n+1} > 2x^{2n}
   \]

4. For the right-hand side, we have:
   \[
   xyz + 2^{2n+1} \leq x(x-2) \cdot 5 \cdot 2^{2n} + 2^{2n+1} = (5x^2 - 10x + 2)2^{2n}
   \]

5. Comparing both sides, we get:
   \[
   2x^{2n} < (5x^2 - 10x + 2)2^{2n}
   \]
   Simplifying, we obtain:
   \[
   2x^{2n} < 5x^2 \cdot 2^{2n} - 10x \cdot 2^{2n} + 2 \cdot 2^{2n}
   \]
   Dividing both sides by \( 2^{2n} \):
   \[
   2x^{2n} < 5x^2 - 10x + 2
   \]

6. To find the range of \( x \), we solve:
   \[
   5x^2 - 10x + 2 > 2 \left( \frac{x}{2} \right)^{2n} \geq 2 \left( \frac{x}{2} \right)^4
   \]
   Simplifying further:
   \[
   5x^2 - 10x + 2 > \frac{x^4}{8}
   \]

7. Solving the inequality \( (x-5)((x-5)(x^2+10x+35)+180) \leq -9 \), we find:
   \[
   x < 5
   \]
   Therefore, \( x \) can be 4, 3, 2, or 1. Since \( x - y \geq 2 \), we only consider \( x = 4 \) and \( x = 3 \).

8. Case 1: \( x = 4 \) and \( y = 2 \)
   \[
   4^{2n+1} - 2^{2n+1} = 8z + 2^{2n+1}
   \]
   Simplifying:
   \[
   4^{2n+1} - 2^{2n+1} \leq 40 \cdot 2^{2n} + 2^{2n+1} = 21 \cdot 2^{2n+1}
   \]
   This implies:
   \[
   2^{2n+1} - 1 \leq 21
   \]
   This is a contradiction since \( 2^{2n+1} - 1 \) grows exponentially.

9. Case 2: \( x = 3 \) and \( y = 1 \)
   \[
   3^{2n+1} - 1 = 3z + 2^{2n+1}
   \]
   Simplifying:
   \[
   3^{2n+1} - 1 \leq 17 \cdot 2^{2n}
   \]
   This implies:
   \[
   \left( \frac{3}{2} \right)^{2n} < 6 < \left( \frac{3}{2} \right)^6
   \]
   Therefore, \( n = 2 \).

10. Substituting \( n = 2 \):
    \[
    3^5 - 1 = 3z + 2^5
    \]
    Simplifying:
    \[
    243 - 1 = 3z + 32
    \]
    \[
    242 = 3z + 32
    \]
    \[
    3z = 210
    \]
    \[
    z = 70
    \]

Thus, the solution is \((x, y, z, n) = (3, 1, 70, 2)\).

The final answer is \( \boxed{ (3, 1, 70, 2) } \).