Let $[x]=a$ and $x=a+b$ where $0\leq b<1$. We have $$(a+b)a+2022=[(a+b)^2]\Rightarrow a^2+ab+2022=a^2+[2ab+b^2]\Rightarrow ab+2022=[2ab+b^2]$$Clearly, RHS is an integer. Hence, $ab$ should be an integer too. Then, $[2ab+b^2]=2ab+[b^2]=2ab$. So the original equation becomes $$ab+2022=2ab\Rightarrow 2022=ab\Rightarrow b=\frac{2022}a$$Since $1>b$, we find that $a\ge 2023$. Clearly, all numbers in the form $k+\frac{2022}k$ works where $k\ge 2023, k\in\mathbb{Z}$. So, we are done.