Sol.

First of all, it's easy to see that both $x$ and $y$ are non-negative. The original equation is equivalent to: $$(\sqrt{x})^2+(2\sqrt{y})^2-343(\sqrt{x}+2\sqrt{y})+2\sqrt{x}\times2\sqrt{y}+2022=0$$ $$\iff (\sqrt{x}+2\sqrt{y})^2-343(\sqrt{x}+2\sqrt{y})+2022=0$$ $$\iff (\sqrt{x}+2\sqrt{y})(\sqrt{x}+2\sqrt{y}-343)+2022=0$$ For convenience, let $\sqrt{x}+2\sqrt{y}=k$. $$\iff k(k-343)+2022=0$$ $$\iff k^2 -343k +2022=0$$ $$\iff (k-6)(k-337)=0$$ Case 1: $k=6$ Then $\sqrt{x}+2\sqrt{y}=6$, $x=(6-2\sqrt{y})^2$, this shows that $y$ must be a perfect square, which also implies that $x$ is a perfect square. Now we know that $\sqrt{x}\ge0$ whenever $0\le\sqrt{y}\le3$, which gives 4 integer solutions to $y$ and it's easy to see that every $y$ has its corresponding $x$. Case 2: $k=337$ Then $\sqrt{x}+2\sqrt{y}=337$. By the same argument we used in case 1, both $x$ and $y$ are perfect squares. So now we have $\sqrt{x}\ge0$ whenever $0\le\sqrt{y}\le\frac{337}{2}$, which gives 169 integer solutions to $y$ and it's corresponding $x$. Thus in total we have $4+169=173$ integer solutions that satisfy this equation.

Let $\sqrt{x}=a$ and $\sqrt{y}=b$, by definition, $a,b>=0$ The equation is $$a^2+4b^2-343a-686b+4ab+2022=(a+2b)^2-343(a+2b)+2022=0$$Therefore, $a+2b=\frac{343\pm 331}{2}$, therefore, we just need to test 2 cases Case 1: $a+2b=6\to \lfloor \frac{6}{2} \rfloor +1=4$ solutions Case 2: $a+2b=337\to \lfloor\frac{337}{2} \rfloor+1=169$ solutions Therefore, there are $169+4=\boxed{173}$ solutions