One. (20 points) Let $P\left(x+a, y_{1}\right)$, $Q\left(x, y_{2}\right)$, and $R\left(2+a, y_{3}\right)$ be three distinct points on the graph of the inverse function of $f(x)=2^{x}+a$, and they satisfy $y_{1}+y_{3}=2 y_{2}$. Find the range of real numbers $a$ for which there is exactly one real number $x$ that satisfies this condition.
$-f(x)=2^{x}+a$ has the inverse function $f^{-1}(x)=\log _{2}(x-a)$.
Then $y_{1}=\log _{2} x, y_{2}=\log _{2}(x-a), y_{3}=1$.
From $y_{1}+y_{3}=2 y_{2}$, we get $1+\log _{2} x=2 \log _{2}(x-a)$. This equation is equivalent to $\left\{\begin{array}{l}x>a, \\ 2 x=(x-a)^{2}\end{array}\right.$.
(1) $\Delta=0$, i.e., $a=-\frac{1}{2}$, the equation has a unique real root $x=\frac{1}{2}$.
(2) $\Delta>0$, i.e., $a>-\frac{1}{2}$, the equation has two real roots $x=$ $a+1 \pm \sqrt{2 a+1}$.
Clearly, $x=a+1+\sqrt{2 a+1}>a$ satisfies the condition.
Thus, we should have $x=a+1-\sqrt{2 a+1} \leqslant a$. Solving this gives $a \geqslant 0$. When $a=0$, points $P, Q$ coincide, which is a contradiction. Therefore, $a>0$.
In summary, the range of real number $a$ is $\left\{-\frac{1}{2}\right\} \cup(0,+\infty)$.