7.265. $\left\{\begin{array}{l}\log _{x} y=2, \\ \log _{x+1}(y+23)=3 .\end{array}\right.$
Solution.

Domain of definition: $\left\{\begin{array}{l}y>0, \\ 0<x \neq 1 .\end{array}\right.$

The system of equations is equivalent to the following:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ y = x ^ { 2 } , } \\
{ y + 23 = ( x + 1 ) ^ { 3 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=x^{2} \\
x^{2}+23=x^{3}+3 x^{2}+3 x+1
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array}{l}
y=x^{2} \\
x^{3}+2 x^{2}+3 x-22=0
\end{array}\right.
\end{aligned}
$$

Divide the left side of the second equation of this system by $x-2$ :

$$
\begin{aligned}
& \begin{array}{l|l}
-x^{3}+2 x^{2}+3 x-22 & x-2 \\
\hline \frac{x^{3}-2 x^{2}}{4 x^{2}}+3 x & x^{2}+4 x+11
\end{array} \\
& \begin{aligned}
\frac{4 x^{2}-8 x}{11 x}-22 \\
-\frac{11 x-22}{0}
\end{aligned}
\end{aligned}
$$

Then the second equation can be represented as $(x-2)\left(x^{2}+4 x+11\right)=0$

from which $x=2$, since $x^{2}+4 x+11 \neq 0(D<0)$. Hence, $y=2^{2}=4$.

Answer: $(2 ; 4)$.