Example 6. For what integer values of $m$ does the equation $4 x^{2}-4(m+1) x-3 m^{2}+10 m-3=0$ have two integer solutions? Two positive integer solutions?
Solve $\because \Delta=16(2 m-2)^{2}$,
$$
\begin{array}{l} 
\therefore x_{1}=\frac{3 m-1}{2}=m+\frac{m-1}{2}, \\
x_{2}=\frac{-m+3}{2}=1-\frac{m-1}{2}, \\
\text { let } \quad m-1=2 k, \quad(k \text { is an integer })
\end{array}
$$

Let $m-1=2 k$, ($k$ is an integer)
Then $m=2 k+1$.
(3) Substituting into (1), (2) gives $x_{1}=3 k+1, x_{2}=1-k$.

Therefore, when $m=2 k+1$, the equation has two integer solutions, with the general form $x_{1}=3 k+1, x_{2}=1-k$.
Since $x>0$, solving gives $-\frac{1}{3}<k<1$.
Since $k$ is an integer, only $k=0$ is valid. Substituting into (3),
(1), (2) shows that when $m=1$, the equation has two positive integer solutions $x_{1}=1, x_{2}=1$.

Summary $3^{\circ}$ When $\Delta=A m^{2}+B m+C$ is a perfect square, the solutions of the equation usually have two cases: (I)
$$
x=\frac{A}{ \pm B m+C} ; \quad(\pi) \quad x= \pm B m+C
$$
($A$, $B$ are rational numbers, $C$ is an integer).
For case (I), use congruence analysis and factor analysis to determine the values of $m$ (as in Example 5).

Case (II) is a linear Diophantine equation in two variables. First, separate the coefficient of $m$ and the constant term into integer and fractional parts (as in Example 6), set the fractional part to be an integer $k$, resulting in another Diophantine equation. Continue to handle it according to the described method until the integer general form of $m$ and $x$ is obtained.