7.273. $\log _{9}\left(x^{3}+y^{3}\right)=\log _{3}\left(x^{2}-y^{2}\right)=\log _{3}(x+y)$.
## Solution.

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

Rewrite the given double equality as a system of equations $\left\{\begin{array}{l}\log _{9}\left(x^{3}+y^{3}\right)=\log _{3}(x+y) \\ \log _{3}\left(x^{2}-y^{2}\right)=\log _{3}(x+y)\end{array}\right.$ In the first equation of the system, switch to base 3. We have

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

Since $x+y>0$, we get

$$
\left\{\begin{array}{l}
x^{2}-x y+y^{2}=x+y \\
x-y=1
\end{array} \Rightarrow x=1+y\right.
$$

Then from the first equation of the system we have

$(1+y)^{2}-(1+y) y+y^{2}=1+y+y, y^{2}-y=0$,

from which $y_{1}=0, y_{2}=1$. Then $x_{1}=1, x_{2}=2$.

Answer: $(1 ; 0),(2 ; 1)$.