One, (20 points) Given the quadratic equation in $x$
$$
\left(k^{2}-6 k+8\right) x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}-4=0
$$
has two integer roots. Find the value of the real number $k$.


The original equation can be transformed into
$$
[(k-4) x+(k-2)][(k-2) x+(k+2)]=0 \text {. }
$$

Since $(k-4)(k-2) \neq 0$, we have
$$
\begin{array}{l}
x_{1}=-\frac{k-2}{k-4}=-1-\frac{2}{k-4}, \\
x_{2}=-\frac{k+2}{k-2}=-1-\frac{4}{k-2} .
\end{array}
$$

Thus, $k-4=-\frac{2}{x_{1}+1}$,
$$
k-2=-\frac{4}{x_{2}+1},
$$

where $x_{1} \neq-1, x_{2} \neq-1$.
Subtracting the two equations and simplifying, we get
$$
x_{1}\left(x_{2}+3\right)=-2 \text {. }
$$

Since $x_{1}$ and $x_{2}$ are both integers, we have
$$
\left(x_{1}, x_{2}\right)=(-2,-2),(1,-5),(2,-4) \text {. }
$$

Therefore, $k=6,3, \frac{10}{3}$.
Upon verification, all satisfy the problem's conditions.