In the Cartesian coordinate system XOY, we color the points with natural number horizontal coordinates and perfect square vertical coordinates (squares of natural numbers) as red points. Try to determine all the "red points" that the graph of the function $y=(x-36)(x-144)-1991$ passes through.
Let $y=m^{2}, m \in N$, then
$$
(x-90)^{2}-4907=m^{2} \text {. }
$$

Let $x-90=k$, then $k^{2}-m^{2}=4907$. Since $7 \times 701$ are both prime numbers, and $k-m<k+m$, we have
$$
\left\{\begin{array}{l}
k-m=1, \\
k+m=4907
\end{array}\right.
$$

which gives $k=2454, m=2453, x=2544$;
or $\left\{\begin{array}{l}k-m=7, \\ k+m=701\end{array}\right.$
which gives $k=354, m=347, x=444$.
When $\left\{\begin{array}{l}k-m=-4907, \\ k+m=-1\end{array}\right.$ and $\left\{\begin{array}{l}k-m=-701, \\ k+m=-7\end{array}\right.$,
the solutions for $x$ are both negative integers, which do not meet the requirements.
Therefore, its graph passes through only two red points, which are $(444,120409)$ and $(2544,6017209)$.