All possible $(x, y, z)$ are $x = y = [1, 2, \ldots, 2^k - 1]$ and $z = [1, 2, \ldots, 2^k]$, where $k$ is any positive integer and $[\ldots]$ denotes the least common multiple.

Clearly $z > x, y$. Suppose $x = [1, \ldots, a]$, $y = [1, \ldots, b]$, and $z = [1, \ldots, c]$, and without loss of generality assume $a \le b$. Furthermore, without loss of generality let $a$ be maximal (that is, for any $a' > a$ we have $x \ne [1, \ldots, a']$), let $b$ be maximal, and let $c$ be minimal. This necessarily guarantees that $c > b$ and $y \mid z - y = x$, so $x = y$ and $z = 2x$. By this maximality and minimality, this means $a = b = c - 1$, and also $2^{v_2(c)} \nmid y$ but $2^{v_2(c)-1} \mid y$. This can only hold when $c$ is some power of $2$ (otherwise $2^{v_2(c)} < c$ would divide $y$.) Hence the solutions must be of the form $x = y = [1, 2, \ldots, 2^k - 1]$ and $z = [1, 2, \ldots, 2^k]$, where $k$ is any positive integer. Substituting back to check shows this clearly holds. Q.E.D.