Three. (25 points) Find the real number $k(k \neq \pm 1)$, such that the equation
$$
\left(k^{2}-1\right) x^{2}-6(3 k-1) x+72=0
$$
has two positive integer roots.
$$
x_{1}=\frac{12}{k+1}, x_{2}=\frac{6}{k-1} \text {. }
$$

Given that $x_{1}, x_{2}$ are positive integers, $k$ must be a rational number greater than 1. Let $k=\frac{m}{n}(m$ and $n$ are coprime, $m>n>0)$. Then
$$
\begin{array}{l}
x_{1}=\frac{12 n}{m+n}, x_{2}=\frac{6 n}{m-n} . \\
\text{Since } (m+n, n)=(m-n, n)=1, \text{ hence } \\
(m+n, m-n) \leqslant 2,
\end{array}
$$

Then $(m+n) \mid 12, (m-n) \mid 6$.
Since $m+n, m-n$ are both even, and
$$
m+n>m-n \text {, }
$$

Therefore, $(m+n, m-n)=(3,1),(4,2),(6,2),(12,2)$.
Solving gives $(m, n)=(2,1),(3,1),(4,2),(7,5)$.
Thus, $k=2,3, \frac{7}{5}$ satisfy the conditions.