Clearly, $(x,y)=(1, 1)$ and $(0, k)$ where $k$ is a non-negative integer are solutions. We are going to show that they are the only solutions.

Let $p_1$ be a prime divisor of $x$. Then because $2^x -1 = xy$, it must be true that $p_1|2^x-1$ . Additionally, let $q$ be the least positive integer such that $p_1 \mid 2^q-1$. From Fermat's Little Theorem, we have $ p_1 |2^{p_1-1} -1$, and therefore $q\le p_1-1 < p_1$. Let's now prove that $q\mid x$. If this is not the case, then $x=kq+r$, with $0 < r < q$, and \begin{align} 2^x -1 &= 2^{kq}2^r - 1\\ &= (2^q)^k \cdot 2^r - 1\\ &= (2^q -1 + 1)^k \cdot 2^r -1\\ &\equiv 2^r - 1\pmod{p_1} \end{align} The last step holds because we have assumed $p_1|2^q-1$, therefore $(2^q-1+1)\equiv 1\pmod{p_1}$. Meanwhile, the relation $2^x -1\equiv 2^r-1\pmod{p_1}$ means $p_1|2^r-1$ because we know $p_1|2^x-1$. However, this contradicts to the minimality of $q$. Consequently, it must be true that $q|x$ and $0 < q < p_1$. Now, let $p_2$ be a prime divisor of $q$. Then, because $q|x$, we must ahve $p_2|x$. Because $q < p_1$, it must hold that $p_2 < p_1$. Therefore, starting with any prime divisor of $x$ (e.g. $p_1$), we can always find a smaller prime $p_2 < p_1$ which divides $x$. This is impossible by the principle of infinite descend. Therefore $x$ cannot have any prime divisor which means $x$ can only be $1$ or $0$.