Example 4 The equation about $x, y$
$$
x^{2}+x y+2 y^{2}=29
$$

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

Since the equation has integer roots, $\Delta \geqslant 0$, and it must be a perfect square.

From $\Delta=y^{2}-4\left(2 y^{2}-29\right)=-7 y^{2}+116 \geqslant 0$, solving for $y^{2} \leqslant \frac{116}{7}<17$.
Thus, the possible values of $y^{2}$ are $0, 1, 4, 9, 16$.
Clearly, only when $y^{2}=16$, $\Delta=4$ is a perfect square, which meets the requirement.

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

When $y=-4$, the original equation becomes $x^{2}-4 x+3=0$, at this time, $x_{1}=1, x_{2}=3$.
Therefore, the answer is C.