Example 6 Let the quadratic equation $\left(k^{2}-6 k+8\right) \cdot x^{2}+\left(2 k^{2}-6 k-4\right) x+k^{2}=4$ have two integer roots. Find all real values of $k$ that satisfy the condition.
Analysis: The expression of the equation is relatively complex, and the discriminant method and Vieta's formulas are not applicable. Transform the original equation to get
$$
\begin{array}{l}
(k-2)(k-4) x^{2}+\left(2 k^{2}-6 k-4\right) x \\
+(k-2)(k+2)=0 .
\end{array}
$$

Factorizing, we get
$$
[(k-2) x+k+2][(k-4) x+k-2]=0 \text {. }
$$

Clearly, $k \neq 2, k \neq 4$. Solving, we get
$$
x_{1} \doteq-\frac{k-2}{k-4}, x_{2}=-\frac{k+2}{k-2} \text {. }
$$

Eliminating $k$, we get
$$
\begin{array}{l}
x_{1} x_{2}+3 x_{2}+2=0 . \\
\therefore x_{2}\left(x_{1}+3\right)=-2 . \\
\text { Discussing, we get }\left\{\begin{array}{l}
x_{2}=-2, \\
x_{1}+3=1 ;
\end{array}\right.
\end{array}
$$

Discussing, we get $\left\{\begin{array}{l}x_{2}=-2, \\ x_{1}+3=1 \text {; }\end{array}\right.$
or $\left\{\begin{array}{l}x_{1}=2, \\ x_{2}+3=-1 ;\end{array}\right.$ or $\left\{\begin{array}{l}x_{1}=1, \\ x_{2}+3=-2 .\end{array}\right.$
Solving, we get $x_{1} 、 x_{2}$, substituting into the original equation, we get the values of $k$ as $6,3, \frac{10}{3}$. Note: When the discriminant and Vieta's formulas are difficult to solve, it often means that the equation can be solved using factorization, and then the results can be transformed and organized.