How small can it get?

$\log_2 x + \log_2 y \ge 6$ $\implies \log_2 xy \ge 6$ $\implies xy \ge 2^6 = 64$ . Since $x, y > 0$ , else $\log_2 x$ and $\log_2 y$ are not defined, we can apply AM-GM inequality as $x+y \ge 2\sqrt{xy} \ge \boxed{16}$ .

$\log_2{x}+\log_2{y}\geq 6,xy\ge64$ ,let $xy=64$ then the minimum of $x+y$ happens when $x=y=8$ ,so the answer is $16$