Example 8 Find the positive integer solutions of the equation
$$
3 x^{2}-8 x y+7 y^{2}-4 x+2 y=109
$$
Solution: The original equation can be transformed to
$$
(3 x-4 y-2)^{2}+5(y-1)^{2}=336 \text {. }
$$

From $5(y-1)^{2}=336-(3 x-4 y-2)^{2} \leqslant 336$,
we get $1 \leqslant y \leqslant 9$.
Moreover, since a perfect square is congruent to 0 or 1 modulo 4, we have that both sides of equation (1) modulo 4 are 0 or
$$
1+(y-1)^{2} \equiv 0(\bmod 4) \text {. }
$$

Since $(y-1)^{2} \equiv 0,1(\bmod 4)$, it follows that
$$
(y-1)^{2} \equiv 0(\bmod 4) \text {. }
$$

Thus, $y$ is odd. Therefore, $y=1,3,5,7,9$.
Upon inspection, only when $y=5,9$, $336-5(y-1)^{2}$
is a perfect square.
When $y=5$, substituting into the equation gives
$$
3 x-4 y-2= \pm 16 \text {. }
$$

It is easy to see that only $3 x-4 y-2=-16$ has integer solutions, in which case, the solution to the equation is $(x, y)=(2,5)$.
When $y=9$, substituting into the equation gives
$$
3 x-4 y-2= \pm 4 \text {. }
$$

It is easy to see that only $3 x-4 y-2=4$ has integer solutions, in which case, the solution to the equation is $(x, y)=(14,9)$.
In summary, the solutions to the equation are
$$
(x, y)=(2,5),(14,9) \text {. }
$$
(To be continued)