Solution:

The answer is (B). First of all, note that $2024=2^{3} \cdot 11 \cdot 23$ must divide $n^{2}$, so we can write $n=\left(2^{2} \cdot 11 \cdot 23\right) \cdot k=1012 k$, with $k$ an integer of modulus less than or equal to $9$, so that $|n|=|1012 k|<10000$. Substituting into the equation we have
$$1012^{2} k^{2}+2 \cdot 1012^{2} k+2024 m=2024,$$
and dividing by $2024$ we get
$$506 k^{2}+1012 k-1=-m.$$
The number of solutions $(n, m)$ is therefore equal to the number of integers $-10<k<10$ such that
$$-10000<506 k^{2}+1012 k-1<10000 ;$$
indeed, once $k$ is chosen, the integer $m$ is uniquely determined. One easily observes that for $k \leq -6$ or $k \geq 4$ the inequality is not satisfied, while it is satisfied for all integers between $-5$ and $3$, which are $9$.