Example 7 The equation in terms of $x, y$
$$
x^{2}+x y+2 y^{2}=29
$$

has ( ) sets of integer solutions $(x, y)$.
(A)2
(B) 3
(C) 4
(D) infinitely many
We can consider the original equation as a quadratic equation in $x$
$$
x^{2}+y x+\left(2 y^{2}-29\right)=0 \text {. }
$$

Since the equation has integer roots, the discriminant $\Delta \geqslant 0$, and it must be a perfect square.
By $\Delta=y^{2}-4\left(2 y^{2}-29\right)$
$$
=-7 y^{2}+116 \geqslant 0 \text {, }
$$

Solving for $y^{2} \leqslant \frac{116}{7} \approx 16.57$.
Thus, $y^{2}=0,1,4,9,16$.
Upon inspection, 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, $x_{3}=1, x_{4}=3$.
Therefore, the original equation has a total of 4 sets of integer solutions.
Hence, the answer is C.