7.259. $\left\{\begin{array}{l}2-\log _{2} y=2 \log _{2}(x+y) \\ \log _{2}(x+y)+\log _{2}\left(x^{2}-x y+y^{2}\right)=1 .\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
2 - \log _{2} y = 2 \log _{2}(x + y) \\
\log _{2}(x + y) + \log _{2}\left(x^{2} - x y + y^{2}\right) = 1 .
\end{array}\right.
\]
Solution.

Domain of definition: $\left\{\begin{array}{l}x+y>0, \\ y>0, \\ x^{2}-x y+y^{2}>0 .\end{array}\right.$

Write the system of equations as

$$
\begin{aligned}
& \left\{\begin{array}{l}
\log _{2} 4-\log _{2} y=\log _{2}(x+y)^{2}, \\
\log _{2}(x+y)+\log _{2}\left(x^{2}-x y+y^{2}\right)=1 .
\end{array} \Leftrightarrow\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ \operatorname { l o g } _ { 2 } \frac { 4 } { y } = \operatorname { l o g } _ { 2 } ( x + y ) ^ { 2 } , } \\
{ \operatorname { l o g } _ { 2 } ( x + y ) ( x ^ { 2 } - x y + y ^ { 2 } ) = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{4}{y}=(x+y)^{2}, \\
(x+y)\left(x^{2}-x y+y^{2}\right)=2
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x + y = \frac { 2 } { \sqrt { y } } , } \\
{ ( x + y ) ( ( x + y ) ^ { 2 } - 3 x y ) = 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=\frac{2}{\sqrt{y}}-y, \\
\frac{2}{\sqrt{y}}\left(\frac{4}{y}-3 y\left(\frac{2}{\sqrt{y}}-y\right)\right)=2 \\
6 y^{3}-14 y \sqrt{y}+8=0 .
\end{array}\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
x=\frac{2}{\sqrt{y}}-y,
\end{array}\right. \\
& \Leftrightarrow
\end{aligned}
$$

Let $y \sqrt{y}=t$. Then the second equation has the form $6 t^{2}-14 t+8=0$,
$3 t^{2}-7 t+4=0 \Rightarrow t_{1}=1, t_{2}=\frac{4}{3}$. Then $y_{1}=1, x_{1}=1 ; y_{2}=\sqrt[3]{\frac{16}{9}}$, $x_{2}=2 \sqrt[3]{\frac{3}{4}}-\sqrt[3]{\frac{16}{9}}=\frac{2 \sqrt[3]{6}}{2}-\frac{2 \sqrt[3]{6}}{3}=\frac{2 \sqrt[3]{6}}{6}=\frac{\sqrt[3]{6}}{3}$.

Answer: $\quad(1 ; 1),\left(\frac{\sqrt[3]{6}}{3} ; \frac{2 \sqrt[3]{6}}{3}\right)$.