Example 6 If $m$ takes any real number not equal to 0 or 1, the parabola
$$
y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}
$$

in the Cartesian coordinate system always passes through two fixed points, then, the distance between these two fixed points is $\qquad$ .
$$
y=\frac{m-1}{m} x^{2}+\frac{2}{m} x-\frac{m-3}{m}
$$

Reversing the roles, rearranging as a linear equation in $m$, we get
$$
\left(x^{2}-y-1\right) m+\left(-x^{2}+2 x+3\right)=0 \text {. }
$$

Given that when $m$ is any real number not equal to 0 or 1, the parabola passes through two fixed points, we have
$$
\left\{\begin{array}{l}
x^{2}-y-1=0, \\
-x^{2}+2 x+3=0 .
\end{array}\right.
$$

Solving, we get $x_{1}=3, y_{1}=8 ; x_{2}=-1, y_{2}=0$.
Therefore, the coordinates of the two fixed points $A$ and $B$ are
$$
A(3,8) 、 B(-1,0) \text {. }
$$

Thus, $A B=\sqrt{[3-(-1)]^{2}+(8-0)^{2}}=4 \sqrt{5}$.
Note: This problem can also be solved by substituting two special values of $m$, thereby obtaining two special parabolas, and finding the coordinates of the intersection points of these two parabolas, which are the required fixed point coordinates.