jmc

By AM-GM, $$1=16^{x^2+y}+16^{x+y^2}\ge 2\sqrt{16^{x^2+y}\cdot 16^{x+y^2}}=2\cdot 4^{x^2+y^2+x+y}=2^{2x^2+2y^2+2x+2y+1},$$so $$2x^2+2y^2+2x+2y+1\le 0.$$Then $$x^2+x+y^2+y+\frac{1}{2}\le 0.$$Completing the square in $x$ and $y,$ we get $${\left(x+\frac{1}{2}\right)}^2+{\left(y+\frac{1}{2}\right)}^2\le 0.$$The only possible pair is then $(x,y)=\left(-\frac{1}{2},-\frac{1}{2}\right).$ Hence, there is only $\boxed{1}$ solution.