Example 4. For what integer values of $m$ does the equation $x^{2}-(m-1) x-2 m^{2}-5 m-3=0$ have integer solutions?

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Solving by imitating Example 3, we get
$$
\left\{\begin{array}{l}
m=-1, \\
k=2 .
\end{array}\right.
$$

Substituting into (2), we get $x=0,-2$.
Therefore, when $m=-1$, the equation has integer solutions.
Summary $2^{\circ}$ When $\Delta=A m^{2}+B m+C$ and
$A>0$, and $A$ is a perfect square, the discriminant can be completed by squaring and setting it equal to $k^{2}$ (where $k$ is a non-negative integer), then factorizing, and through discussion, deriving a system of equations, thereby determining the integer solutions $x$.

It should be noted here: When $\Delta=A m^{2}+B m$ $+C$ and $A$ is not a perfect square, the integer solutions of the equation (if they exist) are not finite integers. Solving them goes beyond the scope of this article, and will not be discussed here. For example, for a certain parametric quadratic equation, the discriminant $\Delta=6 m^{2}+24 m$ $+34=6(m+2)^{2}+10$, if we set $\Delta=k^{2}$, then $[\sqrt{6}(m+2)-k][\sqrt{6}(m+2)+k]$ $=-10$. Since $m, k$ are integers, $\sqrt{6}(m+2)$ $\pm k$ are not integers, so the above method fails.