Find all positive integers $k$ for which there exist positive integers $x, y$, such that $\frac{x^ky}{x^2+y^2}$ is a prime. 
We need to find all positive integers \( k \) for which there exist positive integers \( x \) and \( y \) such that \(\frac{x^k y}{x^2 + y^2}\) is a prime number \( p \).

1. **Case \( k = 1 \):**
   \[
   \frac{xy}{x^2 + y^2} = p
   \]
   This implies:
   \[
   xy = p(x^2 + y^2)
   \]
   Rearranging, we get:
   \[
   xy = px^2 + py^2 \implies xy \geq 2pxy \implies 1 \geq 2p
   \]
   Since \( p \) is a prime number, \( p \geq 2 \), which makes \( 1 \geq 4 \), a contradiction. Therefore, \( k = 1 \) is not possible.

2. **Case \( k \geq 2 \):**
   Let \( d = \gcd(x, y) \), so \( x = dx_1 \) and \( y = dy_1 \) where \(\gcd(x_1, y_1) = 1\). The given condition becomes:
   \[
   \frac{(dx_1)^k (dy_1)}{(dx_1)^2 + (dy_1)^2} = p \implies \frac{d^{k+1} x_1^k y_1}{d^2 (x_1^2 + y_1^2)} = p \implies d^{k-1} x_1^k y_1 = p (x_1^2 + y_1^2)
   \]

3. **Case 1: \( p \mid y_1 \):**
   Let \( y_1 = pm \). Then:
   \[
   d^{k-1} x_1^k m = x_1^2 + p^2 m^2
   \]
   Since \( x_1 \mid x_1^2 + p^2 m^2 \), \( x_1 \mid p^2 m^2 \). Given \(\gcd(x_1, y_1) = 1\), \( x_1 \mid 1 \implies x_1 = 1 \). Thus, \( x = d \) and:
   \[
   d^{k-1} m = 1 + p^2 m^2
   \]
   Since \( m \mid 1 \), \( m = 1 \). Therefore, \( y = dp \) and:
   \[
   d^{k-1} - 1 = p^2
   \]
   - For \( k = 2 \):
     \[
     d - 1 = p^2 \implies d = p^2 + 1
     \]
     Thus, \( x = p^2 + 1 \) and \( y = p(p^2 + 1) \). This is a valid solution for \( k = 2 \).

   - For \( k \geq 3 \):
     \[
     d^{k-1} - 1 = (d-1)(d^{k-2} + d^{k-3} + \cdots + d + 1) = p^2
     \]
     Since \( d-1 \geq 2 \), the left-hand side is larger than \( p^2 \), leading to a contradiction. Therefore, \( k \geq 3 \) is not possible in this case.

4. **Case 2: \( p \mid x_1 \):**
   Let \( x_1 = pn \). Then:
   \[
   d^{k-1} p^k n^k y_1 = p (p^2 n^2 + y_1^2) \implies d^{k-1} p^{k-1} n^k y_1 = p^2 n^2 + y_1^2
   \]
   Since \( k \geq 2 \), \( p \mid y_1^2 \implies p \mid y_1 \), which contradicts \(\gcd(x_1, y_1) = 1\). Therefore, this case is not possible.

5. **Case 3: \( p \mid d^{k-1} \):**
   Let \( d = pd_1 \). Then:
   \[
   p^{k-2} d_1^{k-1} x_1^k y_1 = x_1^2 + y_1^2
   \]
   Since \( y_1 \mid x_1^2 \), let \( y_1 = 1 \). Then:
   \[
   x_1 \mid x_1^2 + 1 \implies x_1 = 1
   \]
   Thus:
   \[
   p^{k-2} d_1^{k-1} = 2
   \]
   - For \( k = 3 \):
     \[
     p d_1^2 = 2 \implies p = 2 \text{ and } d_1 = 1 \implies d = 2
     \]
     Thus, \( x = y = 2 \). This is a valid solution for \( k = 3 \).

   - For \( k \geq 4 \):
     \[
     p^{k-2} d_1^{k-1} = 2
     \]
     This leads to a size contradiction since \( p \geq 2 \) and \( k-2 \geq 2 \). Therefore, \( k \geq 4 \) is not possible.

Conclusion:
The only possible values for \( k \) are \( k = 2 \) and \( k = 3 \).

The final answer is \( \boxed{ k \in \{2, 3\} } \)