We have that $x^2+xy+y^2=7$ and $x^2-xy+y^2=5$. Subtracting the two equations we get that $2xy=2 \implies xy=1$. We also get that $(x+y)^2=7+xy \implies x+y=\pm 2\sqrt{2}$ and that $(x-y)^2=5-xy \implies x-y=\pm 2$. Now we simply must bash through all of our cases: $$xy=1 \text{ and } x-y=2 \implies \boxed{(1-\sqrt{2},-1-\sqrt{2}) \text{ and } (1+\sqrt{2},\sqrt2-1)}$$$$xy=1 \text{ and } x-y=-2 \implies \boxed{(-1-\sqrt{2},1-\sqrt{2}) \text{ and } (\sqrt{2}-1,1+\sqrt{2})}$$There are, however, more solutions, if $x-y=0$ or $x+y=0$, since the LHS will be $0$ as well. So, we can substitute $x$ for $-y$ or $x$ for $y$, which gives us the other solution pairs $\boxed{(-\sqrt{5},-\sqrt{5}) \text{ and } (\sqrt{5},\sqrt{5})}$ as well as $\boxed{(-\sqrt{7},-\sqrt{7}) \text{ and } (\sqrt{7},\sqrt{7})}$. Finally, if both $x+y=0$ and $x-y=0$ we have the final solution pair of $\boxed{(0,0)}$.