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 $b=1, f(x, y)=x y\left(x^{2}+y^{2}\right)=x y(1-2 x y)$, which is maximized at $x y=\frac{1}{4} \Longleftrightarrow x=y=\frac{1}{2}$, so $(3,1)$ works. However, if $a=4$ and $b=1, f(x, y)=x y\left(x^{3}+y^{3}\right)=x y\left((x+y)^{3}-3 x y(x+y)\right)=x y(1-3 x y)$, 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 $(*$ ):