Example 2 The equation about $x$ and $y$
$$
x^{2}+x y+2 y^{2}=29
$$

has ( ) groups of integer solutions $(x, y)$.
(A) 2
(B) 3
(C) 4
(D) infinitely many
(2009, National Junior High School Mathematics Competition)
Solve: Regarding the original equation as a quadratic equation in $x$, it can be transformed into
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$

Since the equation has integer roots, the discriminant $\Delta$ must be a perfect square.
$$
\begin{array}{l}
\text { By } \Delta=y^{2}-4\left(2 y^{2}-29\right) \\
=-7 y^{2}+116 \geqslant 0 \\
\Rightarrow y^{2} \leqslant \frac{116}{7} \Rightarrow y^{2} \leqslant 16 \\
\Rightarrow y^{2}=16,9,4,1,0 .
\end{array}
$$

Upon verification, only when $y^{2}=16$, $\Delta=4$ is a perfect square, which meets the condition.

When $y=4$, the original equation becomes $x^{2}+4 x+3=0$, solving for $x_{1}=-1, x_{2}=-3$;

When $y=-4$, the original equation becomes $x^{2}-4 x+3=0$, solving for $x_{3}=1, x_{4}=3$.
Therefore, the integer solutions of the original equation are
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = - 1 , } \\
{ y _ { 1 } = 4 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = - 3 , } \\
{ y _ { 2 } = 4 ; }
\end{array} \left\{\begin{array}{l}
x_{3}=1, \\
y_{3}=-4 ;
\end{array} ;\left\{\begin{array}{l}
x_{4}=3, \\
y_{4}=-4 .
\end{array}\right.\right.\right.\right.
$$

Therefore, the correct choice is C.