Solution:

Let $f(x, y) = x^{a} y^{b} + y^{a} x^{b}$. Observe that $2^{1-a-b}$ is merely the value of $f\left(\frac{1}{2}, \frac{1}{2}\right)$, so this value is always achievable.

We claim (call this result (*)) that if $(a, b)$ satisfies the condition, so does $(a+1, b+1)$. To see this, observe that if $f(x, y) \leq 2^{1-a-b}$, then multiplying by the inequality $x y \leq \frac{1}{4}$ yields $x^{a+1} y^{b+1} + y^{a+1} x^{b+1} \leq 2^{-1-a-b}$, as desired.

For the rest of the solution, without loss of generality we consider the $a \geq b$ case. If $a = b = 0$, then $f(x, y) = 2$, so $(0,0)$ works. If $a = 1$ and $b = 0$, then $f(x, y) = x + y = 1$, so $(1,0)$ works. For $a \geq 2$, $(a, 0)$ fails since $f(1,0) = 1 > 2^{1-a}$.

If $a = 3$ and $b = 1$, $f(x, y) = x y (x^{2} + y^{2}) = x y (1 - 2 x y)$, which is maximized at $x y = \frac{1}{4} \Longleftrightarrow x = y = \frac{1}{2}$, so $(3,1)$ works. However, if $a = 4$ and $b = 1$, $f(x, y) = x y (x^{3} + y^{3}) = x y ((x+y)^{3} - 3 x y (x+y)) = x y (1 - 3 x y)$, which is maximized at $x y = \frac{1}{6}$. Thus $(4,1)$ does not work.

From these results and $(*)$, we are able to deduce all the pairs that do work ($\swarrow$ represents those pairs that work by (*)):