Find all pairs $(x, y)$ such that $16^{x^{2}+y}+16^{x+y^{2}}=1 $ 
Find all pairs $(x, y)$ of real numbers such that
$$
16^{x^{2}+y}+16^{x+y^{2}}=1
$$

I know this question has been answered multiple times but i have another solution which i have  to validate.
So after taking $16^x16^y$ common from the equation we get
$$16^{x+y}=\dfrac{1}{16^{x(x-1)}+16^{y(y-1)}}$$
Now LHS is increasing while RHS is decreasing so equation can have at most one real root which can be find by putting x=y to get $(x, y)=(-1 / 2,-1 / 2)$ .
Anything wrong ?

I think your reasoning is wrong.
Why the right side decreases?
How you define decreasing for the function of two variables?
By AM-GM  $$1\geq2\sqrt{16^{x^2+y}\cdot16^{x+y^2}}=4^{x^2+y^2+x+y+\frac{1}{2}}=\sqrt2^{(2x+1)^2+(2y+1)^2}\geq1,$$
which gives $$x=y=-\frac{1}{2}.$$
