Example 5 Given the quadratic equation in $x$
$$
\left(k^{2}-8 k+15\right) x^{2}-2(13-3 k) x+8=0
$$

both roots are integers. Find the value of the real number $k$.
Analysis: Since $k$ is a real number, we cannot solve it using the discriminant. We can first find the two roots of the equation $x_{1}=$ $\frac{2}{5-k}, x_{2}=\frac{4}{3-k}$. Since $x_{1} 、 x_{2}$ are integers, we can eliminate $k$ from the above two equations to obtain an indeterminate equation about $x_{1} 、 x_{2}$. Solving for $x_{1} 、 x_{2}$, we can then find $k$.
Solution: The original equation is
$$
(k-3)(k-5) x^{2}-2(13-3 k)+8=0 \text {, }
$$

which is $[(k-5) x+2][(k-3) x+4]=0$.
Solving, we get $x_{1}=\frac{2}{5-k}, x_{2}=\frac{4}{3-k}$.
Therefore, $k=5-\frac{2}{x_{1}}, k=3-\frac{4}{x_{2}}$.
Eliminating $k$, we get $\frac{2}{x_{1}}-\frac{4}{x_{2}}=2$, which is $\left(x_{1}-1\right)\left(x_{2}+2\right)=-2$.
Thus, $\left\{\begin{array}{l}x_{1}-1=-2,-1,1,2, \\ x_{2}+2=1,2,-2,-1 .\end{array}\right.$
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = 2 , } \\
{ x _ { 2 } = - 4 ; }
\end{array} \left\{\begin{array}{l}
x_{1}=3, \\
x_{2}=-3 .
\end{array}\right.\right.
$$

Therefore, $k=5-\frac{2}{x_{1}}=7,4, \frac{13}{3}$.
Hence, the required $k=4,7, \frac{13}{3}$.