solve for real $x,y$
$$2^{x^2+y}+2^{x+y^2}=8 \tag{1}$$
$$\sqrt{x}+\sqrt{y}=2 \tag{2}$$
Trivially $x=y=1$ Now Equation $(1)$ can be written as
$$2^{x^2+(\sqrt{y})^2}+2^{x+(\sqrt{y})^4}=8$$ so we get
$$2^{x^2+(2-\sqrt{x})^2}+2^{x+(2-\sqrt{x})^4}=8$$ so
$$2^{x^2+4+x-4\sqrt{x}}+2^{x^2+25x+16-32\sqrt{x}-8x\sqrt{x}}=8$$
But i have no clue to proceed..
We will show that the solution $(x,y)=(1,1)$ is unique. $$2^{x^2+y}+2^{x+y^2}=8 \tag{1}\\ln(2^{x^2+y}+2^{x+y^2})=ln8⇒$$ We assume that $x^2+y>y^2+x$, (similarly you can assume that $x^2+y>y^2+x$ and reach the same result) and it follows that: $$ln(2^{x^2+y}+2^{x+y^2})=\Big(ln2^{x^2+y}\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)\Big)=(x^2+y)ln2+ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)$$ And since $x^2+y>y^2+x$, we have: $${\frac{2^{y^2+x}}{2^{x^2+y}}}<1$$ So, $$ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)<ln2⇒\\(x^2+y)ln2+ln\Big(1+\frac{2^{y^2+x}}{2^{x^2+y}}\Big)=ln8<(x^2+y)ln2+ln2⇒\\(x^2+y)+1>\frac{ln8}{ln2}=3$$ And we get, $$x^2+y>2$$ and our second equation which we did not use till now, $$\sqrt{x}+\sqrt{y}=2$$ You can show that these two lead to the solutions $(0,4)$ and $(4,0)$, which do not satisfy our initial condition.
Therefore it must hold that:$$x^2+y=y^2+x\\\sqrt{x}+\sqrt{y}=2$$ Solving this last system will involve a cubic polynomial with one real and two complex solutions. The real solution leads to $(x,y)=(1,1)$