Solve $y$ with respect to $x$ leads to $$x=\frac{7-2y}{2-y}=2+\frac{3}{2-y}$$ This implies that $3$ is divisible by $(2-y)$ which means $y=-1, 5, 1, 3$. Accordingly $$(x,y = (-1, 3), (5, 1), (1, 5), (3, -1)$$