Example 1 Given a quadratic equation $x^{2} + m x - m + 1 = 0$ ($m$ is an integer) has two distinct positive integer roots. Find the value of $m$.
(1996, Sichuan Province Math Competition)
Analysis: For a quadratic equation to have integer roots, it must first satisfy that the discriminant is a perfect square. If we follow this line of thought to calculate and discuss, it will either be very complicated or lead to confusion. However, if we first set the two integer roots, it is easy to derive the form $(\alpha - 1)(\beta - 1)$, which can expedite the problem-solving process and achieve twice the result with half the effort.
Solution: Let the two positive integer roots of the equation be $\alpha, \beta$, then
$$
\begin{array}{l}
\alpha+\beta=-m, \alpha \beta=-m+1 . \\
\therefore \alpha \beta-\alpha-\beta=1 .
\end{array}
$$

Thus, $\alpha \beta-\alpha-\beta+1=2$,
which means $(\alpha-1)(\beta-1)=2$.
Therefore, we have
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ \alpha - 1 = 2 , } \\
{ \beta - 1 = 1 ; }
\end{array} \quad \left\{\begin{array}{l}
\alpha-1=-2, \\
\beta-1=-1 ;
\end{array}\right.\right. \\
\left\{\begin{array} { l } 
{ \alpha - 1 = 1 , } \\
{ \beta - 1 = 2 ; }
\end{array} \quad \left\{\begin{array}{l}
\alpha-1=-1, \\
\beta-1=-2 .
\end{array}\right.\right.
\end{array}
$$

Since $\alpha, \beta$ are positive integers, the two positive integer roots of the equation can only be 3,2.
$$
\therefore m=-5 \text {. }
$$