Determine the positive integers expressible in the form $\frac{x^2+y}{xy+1}$, for at least $2$ pairs $(x,y)$ of positive integers
1. **Auxiliary Result:**
   We need to determine all ordered pairs of positive integers \((x, y)\) such that \(xy + 1\) divides \(x^2 + y\). Suppose \(\frac{x^2 + y}{xy + 1} = k\) is an integer. This implies:
   \[
   xy + 1 \mid x^2 + y \implies xy + 1 \mid x^2y^2 + y^3 = (x^2y^2 - 1) + (y^3 + 1) \implies xy + 1 \mid y^3 + 1.
   \]
   Therefore, we need to find divisors of \(y^3 + 1\) that are congruent to \(1 \pmod{y}\).

2. **Trivial Factors:**
   The trivial factors of \(y^3 + 1\) are \(1\) and \(y^3 + 1\). The factor \(1\) gives no solution. The factor \(y^3 + 1\) leads to:
   \[
   xy + 1 = y^3 + 1 \implies x = y^2.
   \]
   Thus, \((x, y) = (y^2, y)\) is a solution for all \(y \in \mathbb{N}\).

3. **Nontrivial Factors:**
   Suppose \(xy + 1\) is a nontrivial factor of \(y^3 + 1\). Since \(xy + 1\) is congruent to \(1 \pmod{y}\), we can assume without loss of generality that:
   \[
   xy + 1 \leq \sqrt{y^3 + 1} < y^{3/2} + 1 \implies x < y^{1/2} \implies x^2 < y.
   \]
   Then:
   \[
   k = \frac{x^2 + y}{xy + 1} \leq \frac{2y}{xy + 1} \implies kxy + k < 2y.
   \]
   If \(k \geq 2\), then:
   \[
   kxy + k \geq 2y + 2 > 2y,
   \]
   which contradicts the previous inequality. Therefore, \(k = 1\), and we have:
   \[
   x^2 + y = xy + 1 \implies x^2 - xy + y - 1 = 0 \implies (x - 1)(x - y + 1) = 0.
   \]
   This gives us two cases:
   - \(x = 1\)
   - \(x = y - 1\)

4. **Conclusion of Auxiliary Result:**
   The ordered pairs \((x, y)\) satisfying \(xy + 1 \mid x^2 + y\) are:
   \[
   (a^2, a), (1, a), (a - 1, a) \quad \text{for} \quad a \in \mathbb{N}.
   \]

5. **Main Problem:**
   We need to determine the positive integers expressible in the form \(\frac{x^2 + y}{xy + 1}\) for at least two pairs \((x, y)\) of positive integers. From the auxiliary result, we have:
   \[
   \left\{\begin{array}{lc}
   \text{For} \ (x, y) = (a^2, a), & \frac{x^2 + y}{xy + 1} = a, \\
   \text{For} \ (x, y) = (1, a), & \frac{x^2 + y}{xy + 1} = 1, \\
   \text{For} \ (x, y) = (a - 1, a), & \frac{x^2 + y}{xy + 1} = 1.
   \end{array}\right.
   \]
   If \(a \neq 1\), then \(\frac{x^2 + y}{xy + 1}\) takes the value \(a\) only once for \((x, y) = (a^2, a)\). However, it takes the value \(1\) for infinitely many pairs \((x, y) = (1, a)\) and \((x, y) = (a - 1, a)\).

The final answer is \(\boxed{1}\).