Find the number of ordered pairs $(x,y)$ of real numbers such that
\[16^{x^2 + y} + 16^{x + y^2} = 1.\]
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.