Find all natural numbers $ x,y$ such that $ y| (x^{2}+1)$ and $ x^{2}| (y^{3}+1)$.
To find all natural numbers \( x \) and \( y \) such that \( y \mid (x^2 + 1) \) and \( x^2 \mid (y^3 + 1) \), we start by analyzing the given conditions.

1. **Condition 1: \( y \mid (x^2 + 1) \)**

   This implies there exists some integer \( k \) such that:
   \[
   x^2 + 1 = ky
   \]
   Rearranging, we get:
   \[
   x^2 = ky - 1
   \]

2. **Condition 2: \( x^2 \mid (y^3 + 1) \)**

   This implies there exists some integer \( j \) such that:
   \[
   y^3 + 1 = jx^2
   \]
   Substituting \( x^2 = ky - 1 \) from Condition 1 into this equation, we get:
   \[
   y^3 + 1 = j(ky - 1)
   \]
   Simplifying, we have:
   \[
   y^3 + 1 = jky - j
   \]
   Rearranging, we get:
   \[
   y^3 + j = jky - 1
   \]
   \[
   y^3 + j + 1 = jky
   \]

3. **Combining the equations:**

   From the above, we have:
   \[
   y^3 + j + 1 = jky
   \]
   Rearranging, we get:
   \[
   y^3 + j + 1 = jky
   \]
   \[
   y^3 = jky - j - 1
   \]
   \[
   y^3 = y(jky - j - 1)
   \]
   \[
   y^2 = jk - j - 1
   \]

4. **Finding specific values:**

   Let's test small values of \( y \) to find natural number solutions.

   - For \( y = 1 \):
     \[
     x^2 + 1 = k \cdot 1 \implies x^2 = k - 1
     \]
     \[
     x^2 \mid (1^3 + 1) \implies x^2 \mid 2
     \]
     The only natural number \( x \) that satisfies \( x^2 \mid 2 \) is \( x = 1 \).

     So, \( (x, y) = (1, 1) \) is a solution.

   - For \( y = 2 \):
     \[
     x^2 + 1 = k \cdot 2 \implies x^2 = 2k - 1
     \]
     \[
     x^2 \mid (2^3 + 1) \implies x^2 \mid 9
     \]
     The natural numbers \( x \) that satisfy \( x^2 \mid 9 \) are \( x = 1 \) and \( x = 3 \).

     - For \( x = 1 \):
       \[
       1^2 + 1 = 2 \implies k = 1
       \]
       This does not satisfy \( x^2 = 2k - 1 \).

     - For \( x = 3 \):
       \[
       3^2 + 1 = 10 \implies k = 5
       \]
       This does not satisfy \( x^2 = 2k - 1 \).

   - For \( y = 3 \):
     \[
     x^2 + 1 = k \cdot 3 \implies x^2 = 3k - 1
     \]
     \[
     x^2 \mid (3^3 + 1) \implies x^2 \mid 28
     \]
     The natural numbers \( x \) that satisfy \( x^2 \mid 28 \) are \( x = 1 \) and \( x = 2 \).

     - For \( x = 1 \):
       \[
       1^2 + 1 = 2 \implies k = 1
       \]
       This does not satisfy \( x^2 = 3k - 1 \).

     - For \( x = 2 \):
       \[
       2^2 + 1 = 5 \implies k = 2
       \]
       This does not satisfy \( x^2 = 3k - 1 \).

   Testing further values of \( y \) will follow a similar pattern, and it becomes clear that the only solution in natural numbers is \( (x, y) = (1, 1) \).

The final answer is \( \boxed{ (x, y) = (1, 1) } \)