Determine whether there exist an infinite number of  positive integers  $x,y $ satisfying the condition: $x^2+y \mid x+y^2.$  Please prove it.
To determine whether there exist an infinite number of positive integers \( x \) and \( y \) satisfying the condition \( x^2 + y \mid x + y^2 \), we will proceed as follows:

1. **Assume \( y = kx \) where \( k \) is a positive integer:**
   \[
   x^2 + kx \mid x + (kx)^2
   \]
   This simplifies to:
   \[
   x^2 + kx \mid x + k^2x^2
   \]

2. **Rewrite the divisibility condition:**
   \[
   x^2 + kx \mid k^2x^2 + x
   \]
   We can factor out \( x \) from the right-hand side:
   \[
   x^2 + kx \mid x(k^2x + 1)
   \]

3. **Simplify the divisibility condition:**
   Since \( x^2 + kx \) divides \( x(k^2x + 1) \), it must also divide any linear combination of \( x^2 + kx \) and \( x(k^2x + 1) \). In particular, it must divide:
   \[
   x(k^2x + 1) - kx(x^2 + kx) = xk^2x + x - kx^3 - k^2x^2 = x - kx^3
   \]
   Therefore:
   \[
   x^2 + kx \mid x - kx^3
   \]

4. **Further simplification:**
   Since \( x^2 + kx \mid x - kx^3 \), we can write:
   \[
   x^2 + kx \mid x^3 + 1
   \]

5. **Find a suitable \( k \):**
   To satisfy the condition \( x^2 + kx \mid x^3 + 1 \), we can choose \( k = x^3 - x + 1 \). This choice ensures that:
   \[
   x^2 + (x^3 - x + 1)x = x^2 + x^4 - x^2 + x = x^4 + x
   \]
   which divides \( x^3 + 1 \).

6. **Verify the solution:**
   For \( k = x^3 - x + 1 \), we have:
   \[
   y = kx = (x^3 - x + 1)x = x^4 - x^2 + x
   \]
   Thus, the pair \( (x, y) = (a, a(a^3 - a + 1)) \) works for every integer \( a \geq 2 \). When \( a = 1 \), \( x \) and \( y \) are not distinct, but for \( a \geq 2 \), \( x \) and \( y \) are distinct.

Conclusion:
There are infinitely many solutions for \( x \) and \( y \) satisfying the given condition.

The final answer is \( \boxed{ (x, y) = (a, a(a^3 - a + 1)) } \) for \( a \geq 2 \).