For some positive integers $m,n$, the system $x+y^2 = m$ and $x^2+y = n$ has exactly one integral solution $(x,y)$. Determine all possible values of $m-n$.
1. We start with the given system of equations:
   \[
   x + y^2 = m \quad \text{(1)}
   \]
   \[
   x^2 + y = n \quad \text{(2)}
   \]

2. Subtract equation (2) from equation (1):
   \[
   x + y^2 - (x^2 + y) = m - n
   \]
   Simplifying, we get:
   \[
   x + y^2 - x^2 - y = m - n
   \]
   \[
   (x - y)(1 - x - y) = m - n
   \]

3. Let \( d_1 = x - y \) and \( d_2 = 1 - x - y \). Then:
   \[
   d_1 d_2 = m - n
   \]

4. Solving for \( y \) in terms of \( d_1 \) and \( d_2 \):
   \[
   y = \frac{1 - d_1 - d_2}{2}
   \]
   For \( y \) to be an integer, \( 1 - d_1 - d_2 \) must be even, implying \( d_1 + d_2 \) is odd.

5. Since \( d_1 d_2 = m - n \), and \( d_1 + d_2 \) is odd, \( m - n \) must be even. This is because the product of two numbers where one is even and the other is odd is even.

6. Given that the system has exactly one integral solution \((x, y)\), there must be a unique pair \((d_1, d_2)\). If \( m - n \) had more than one pair of divisors, it would lead to multiple solutions for \((x, y)\).

7. For \( m - n \) to have exactly one pair of divisors, it must be a power of 2. This is because powers of 2 have a unique pair of divisors: \( (2^k, 1) \) or \( (-2^k, -1) \).

8. Therefore, \( m - n = \pm 2^k \) where \( k \in \mathbb{Z}^+ \).

The final answer is \( \boxed{ \pm 2^k } \) where \( k \in \mathbb{Z}^+ \).