Is there a number that can be represented as $A^B=B^A$ after $16$ , where A and B are 2 distinct real numbers? [duplicate]

Yes there are such numbers.
By this question, we know that if $x^y=y^x$ and $x\ne y,$ then $x=a^{1/(a-1)},y=a^{a/(a-1)}$ for some $a\not=1.$
Then we see that $$y\ln x=a^{a/(a-1)}\ln a/(a-1)\gt\ln a\frac{a}{a-1}\gt\ln a,$$ when $a\gt1.$ So if we choose $a\gt16,$ then $x^y=y^x$ so produced will satisfy the requirement.

For example, taking $a=100,$ we see $x=100^{1/99}\cong 1.047615, y=100^{100/99}\cong104.761575$ and $x^y=y^x\cong130.726237\gt16.$

Hope this helps.

Yes. The power can be larger than $16$.

Consider the set of points on the plane satisfying the equation

$$x^y=y^x$$

You see that those points form a union of two curves: the line $x=y$ and the more interesting part.

By implicit differentiation we get that on that other curve we have $$ \frac{dy}{dx}=\frac{y(x\ln y-y)}{x(y\ln x-x)}. $$

Using that and logarithmic differentiation we get that on that curve

$$ \frac d{dx}\,y^x=y^x\left[\ln y+\frac{x\ln y-y}{y\ln x-x}\right]. $$

Plugging in $x=4, y=2$ shows that at the point $(x,y)=(4,2)$ we arrive at

$$\frac{d(y^x)}{dx}=\frac{16 (1-\ln2\ln4)}{2-\ln4}\approx1.019.$$

Therefore when $x$ is a little bit larger than $4$ the power $y^x=x^y$ will be a little bit larger than $16$.

I hazard a guess that along that curve $y^x=x^y$ attains a local minimum $e^e\approx 15.15$ at the point $x=y=e$. The formula for the derivative breaks down there (which may explain why Mathematica left a tiny gap there in the above figure).