For positive integers $x,y$, find all pairs $(x,y)$ such that $x^2y + x$ is a multiple of $xy^2 + 7$.
1. **Initial Analysis:**
   We need to find all pairs \((x, y)\) such that \(x^2 y + x\) is a multiple of \(xy^2 + 7\). This can be written as:
   \[
   xy^2 + 7 \mid x^2 y + x
   \]

2. **Considering the Case \(7 \nmid x\):**
   If \(7 \nmid x\), then \(\gcd(x, xy^2 + 7) = 1\). This implies:
   \[
   xy^2 + 7 \mid x(xy + 1)
   \]
   Since \(\gcd(x, xy^2 + 7) = 1\), it must also be true that:
   \[
   xy^2 + 7 \mid xy + 1
   \]
   However, this is impossible because \(xy^2 + 7 > xy + 1\). Therefore, \(7 \mid x\).

3. **Let \(x = 7k\):**
   Substituting \(x = 7k\) into the original condition, we get:
   \[
   7ky^2 + 7 \mid 49ky + 7k
   \]
   Simplifying, we have:
   \[
   ky^2 + 1 \mid 7ky + k
   \]
   Since \(\gcd(k, ky^2 + 1) = 1\), it follows that:
   \[
   ky^2 + 1 \mid 7y + 1
   \]
   This implies \(y \leq 7\).

4. **Testing Values of \(y\):**
   - For \(y = 1\):
     \[
     7k + 7 \mid 49k^2 + 7k \implies k + 1 \mid 7k^2 + k \implies k + 1 \mid 7k + 1
     \]
     Simplifying further:
     \[
     k + 1 \mid 7k + 1 - 7k - 7 = -6
     \]
     Thus:
     \[
     k + 1 \mid 6 \implies k \in \{1, 2, 5\}
     \]
     Therefore, the solutions are \((x, y) = (7, 1), (14, 1), (35, 1)\).

   - For \(y = 2\):
     \[
     28k + 7 \mid 98k^2 + 7k \implies 4k + 1 \mid 14k^2 + k \implies 4k + 1 \mid 14k + 1
     \]
     Simplifying further:
     \[
     4k + 1 \mid 14k + 1 - 14k = 1
     \]
     Thus:
     \[
     4k + 1 \mid 2k - 2 \implies k = 1
     \]
     Therefore, the solution is \((x, y) = (7, 2)\).

   - For \(y = 3, 4, 5, 6\):
     It is easy to check that no solutions exist.

   - For \(y = 7\):
     \[
     49k + 1 \mid k(49k + 1)
     \]
     This is true for all \(k\). Therefore, the solutions are of the form \((x, y) = (7k, 7)\).

5. **Conclusion:**
   Combining all the solutions, we have:
   \[
   \boxed{(x, y) = (7, 1), (14, 1), (35, 1), (7, 2), (7k, 7)}
   \]