1. Solve the equation $2^{x}+2^{y}=2^{x y-1}$ in integers.
Solution. Let $x=y$ first. Then the equation becomes $2^{x}+2^{x}=2^{x^{2}-1}$.

From this, $2^{x+1}=2^{x^{2}-1} \Leftrightarrow x+1=x^{2}-1 \Leftrightarrow x=-1$ or $x=2$. Now let the numbers $x$ and $y$ be different. We can assume that $x<y$. Let $y=x+n, n \in \mathbb{N}$. Then $2^{x}+2^{x+n}=2^{x y-1} \Leftrightarrow 1+2^{n}=2^{x y-1-x}$, which is impossible, since the left side is an odd number greater than 2, while the right side is either even (if $x y-1-x \geq 1$) or does not exceed 2 (if $x y-1-x<1$).

Answer: $(-1,-1),(2,2)$.