7.266. $\left\{\begin{array}{l}\left(x^{2}+y\right) 2^{y-x^{2}}=1 \\ 9\left(x^{2}+y\right)=6^{x^{2}-y}\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\left(x^{2}+y\right) 2^{y-x^{2}}=1 \\ 
9\left(x^{2}+y\right)=6^{x^{2}-y}
\end{array}\right.
\]
## Solution.

From the first equation, we have $x^{2}+y=2^{x^{2}-y}$. Substituting this value into the second equation of the system, we get $9 \cdot 2^{x^{2}-y}=6^{x^{2}-y} \Leftrightarrow$ $\Leftrightarrow 3^{2}=3^{x^{2}-y} \Leftrightarrow x^{2}-y=2$. Then $x^{2}+y=2^{x^{2}-y}=2^{2}=4$ and the original system of equations becomes $\left\{\begin{array}{l}x^{2}+y=4, \\ x^{2}-2=2\end{array} \Leftrightarrow\left\{\begin{array}{l}2 x^{2}=6, \\ 2 y=2,\end{array}\left\{\begin{array}{l}x^{2}=3, \\ y=1,\end{array}\right.\right.\right.$ from which $\left\{\begin{array}{l}x_{1}=\sqrt{3}, \\ y_{1}=1 ;\end{array}\left\{\begin{array}{l}x_{2}=-\sqrt{3} \\ y_{2}=1\end{array}\right.\right.$

Answer: $(\sqrt{3} ; 1),(-\sqrt{3} ; 1)$