Solve the following system of equations over the real numbers:

$$
\begin{aligned}
\log _{2} x & =y-3, \\
x^{y} & =2^{10}
\end{aligned}
$$
From the first equation, we get that $x=2^{y-3}$. Substituting this into the second equation and using the monotonicity of the exponential function, we obtain a quadratic equation for $y$. There are two solutions: $x_{1}=1, y_{1}=5$ and $x_{2}=\frac{1}{32}, y_{2}=-2$.