I think there are no solution for this problem.

There are: $(10, 3)$ and $(-10, -3)$. Solution. Rewrite the equation into $(x-y)^2+(2x-y)^2+5y^2=383$. So $y^2 \leq \frac{383}{5} \implies |y| \leq 8 \implies y \in [-8,8]$. Now, rewrite the equation as $5x^2-6xy+7y^2-383=0$ and solve the quadratic equation for $x$. The discriminant is $\Delta=7660-104y^2$, and it should be a perfect square. So, suppose that $a^2=7660-104y^2$. Use $\mod 3$ and you get $y \equiv 0 \pmod 3$. But we had $y \in [-8,8]$, so $y=-6,-3,3$, or $6$. You can easily check the answers and see that $y=3$ and $y=-3$ satisfy the equation.

multiplying with 5 gives us $(5x-3y)^2+26y^2=1915$ then try $y=1,2,\cdots8$ gives us solution : $(10,3),(-10,-3)$

解: 原方程配方, 得 $$(5x-3y)^2+26y^2=1915. \qquad (1)$$因此 $$26y^2\leqslant 1915. $$$$y^2\leqslant 73\frac{17}{26}. \qquad (2)$$ 另一方面, 由$(1)$式, 知$5x-3y$是奇数, 故对$(1)$式模8, 得 $$1+2y^2\equiv 3\pmod{8}. $$而$y^2\equiv 0, 1, 4(\bmod{8})$, 仅有$y^2\equiv 1(\bmod{8})$满足上式. 从而$y$是奇数. 对原方程式模3, 得 $$-x^2+y^2\equiv 2\pmod{3}. $$而完全平方数模3余数只能是0或1, 所以 $$\begin{aligned} x^2&\equiv 1\pmod{3}, \\ y^2&\equiv 0\pmod{3}. \end{aligned}$$故$y$是3的倍数. 结合$(2)$式(注意$y$是奇数), 知 $$y=\pm 3. $$ 所以 $$\begin{cases} y=3, \\ (5x-9)^2=1915-26\times 3^2=41^2; \end{cases}$$或 $$\begin{cases} y=-3, \\ (5x+9)^2=1915-26\times 3^2=41^2. \end{cases}$$ 由模5, 可知 $$\begin{cases} y=3, \\ 5x-9=41; \end{cases}$$或 $$\begin{cases} y=-3, \\ 5x+9=-41. \end{cases}$$解得 $$(x, y)=(10, 3), (-10, -3). $$