One, (20 points) Given that $m$ and $n$ are integers, point $P\left(m, n^{2}\right)$
lies on the graph of the function
$$
f(x)=\left(x^{2}-4\right)\left(x^{2}+12 x+32\right)+4
$$

Find the number of points $P$ that satisfy the condition.
Notice,
$$
\begin{aligned}
n^{2} & =\left(m^{2}-4\right)\left(m^{2}+12 m+32\right)+4 \\
& =(m+2)(m-2)(m+4)(m+8)+4 \\
& =\left(m^{2}+6 m+8\right)\left(m^{2}+6 m-16\right)+4 .
\end{aligned}
$$

Let $t=m^{2}+6 m-4$.
Then $n^{2}=(t+12)(t-12)+4$, which means
$$
(t+n)(t-n)=140 \text {. }
$$

Assume $n$ is a non-negative integer. Clearly, $t+n$ and $t-n$ have the same parity.
Since $t+n \geqslant t-n$, we have
$$
t+n=-2,-10,14,70 \text {. }
$$

Correspondingly, $t-n=-70,-14,10,2$.
Thus, $(t, n)$
$$
=(-36,34),(-12,2),(12,2),(36,34) \text {. }
$$

Therefore, $m^{2}+6 m-4= \pm 36, \pm 12$.
Solving these, we get $m=4,-10,-8,2,-2,-4$.
Hence, there are 6 points $P$ that satisfy the conditions.