Example 1 Let $N=23x+92y$ be a perfect square, and $N$ does not exceed 2392. Then the number of all positive integer pairs $(x, y)$ that satisfy the above conditions is $\qquad$ pairs.
Solution: Since $N=23(x+4 y)$, and 23 is a prime number, there exists a positive integer $k$, such that $x+4 y=23 k^{2}$.
Also, because $23(x+4 y)=N \leqslant 2392$, we have $x+4 y \leqslant 104$.
Therefore, $23 k^{2}=x+4 y \leqslant 104$, which means $k^{2} \leqslant 4$.
Thus, $k^{2}=1,4$.
When $k^{2}=1$, $x+4 y=23$, in this case, $y \leqslant 5$, we can get 5 solutions;

When $k^{2}=4$, $x+4 y=92$, in this case, $y \leqslant 22$, we can get 22 solutions.

In summary, there are a total of $5+22=27$ pairs of positive integers $(x, y)$ that satisfy the conditions.