Chọn D

Ta có 

$\begin{array}{l}{2022}^x-{2022}^{\sqrt{2x+1}}=1-x^2+2\sqrt{2x+1}\\ \iff {2022}^x+x^2+2x+1=2x+2+2\sqrt{2x+1}+{2022}^{\sqrt{2x+1}}\\ \iff {2022}^x+{\left(x+1\right)}^2={\left(\sqrt{2x+1}+1\right)}^2+{2022}^{\sqrt{2x+1}}.\end{array}$

Xét hàm số $f\left(t\right)={2022}^t+{\left(t+1\right)}^2, t\in \left[0;+\infty \right)$

Ta có $f^{\text{'}}\left(t\right)={2022}^t\ln 2022+2\left(t+1\right)>0,\forall t\in \left[0;+\infty \right)$ nên hàm số y = f(t) đồng biến trên khoảng $\left[0;+\infty \right)$.

Khi đó $f\left(x\right)=f\left(\sqrt{2x+1}\right)\iff x=\sqrt{2x+1}\iff \left\{\begin{array}{l}x\ge 0\\ x^2-2x-1=0\end{array}\right.\iff x=1+\sqrt{2}$.

Suy ra a = 1 và b = 2.

Vậy $a+b^3=1+2^3=9$.