Example 1 Find the positive integer solutions of the equation $x^{2}+y^{2}=328$.
Obviously, $x \neq y$. Without loss of generality, assume $x>y>0$. Since 328 is an even number, the parity of $x$ and $y$ must be the same. Therefore, $x \pm y$ is even.

Let $x+y=2 u_{1}, x-y=2 v_{1}$, then $u_{1}, v_{1} \in \mathbf{Z}$, and $u_{1}>v_{1}>0$. Thus,
$$
x=u_{1}+v_{1}, y=u_{1}-v_{1}.
$$

Substituting into the original equation, we get
$$
u_{1}^{2}+v_{1}^{2}=164.
$$

Similarly, let $u_{1}+v_{1}=2 u_{2}, u_{1}-v_{1}=2 v_{2} (u_{2}, v_{2} \in \mathbf{Z}$, and $u_{2}>v_{2}>0)$. Then, we have
$$
u_{2}^{2}+v_{2}^{2}=82.
$$

Again, let $u_{2}+v_{2}=2 u_{3}, u_{2}-v_{2}=2 v_{3}$, we get
$$
u_{3}^{2}+v_{3}^{2}=41.
$$

At this point, $u_{3}$ and $v_{3}$ must be one odd and one even, and $0<v_{3}<u_{3} \leqslant[\sqrt{41}]=6$.

Taking $v_{3}=1,2,3,4,5$, we get the corresponding $u_{3}^{2}=40,37,32,25,16$.
Therefore, it can only be $u_{3}=5, v_{3}=4$.
Thus, $x=18, y=2$.
Considering the symmetry of the equation, the equation has two solutions $(18,2),(2,18)$.