This is a part of Brazil MO 2021/3. Take $A = \varphi^2$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. Then, it can be seen that $\varphi^{2n}+\varphi^{-2n} \in \mathbb{Z}$ and in particular $0 < \varphi^{-2n} < 1$ means that $\lceil A^n \rceil = \varphi^{2n}+\varphi^{-2n} = (\varphi^n+\varphi^{-n})^2-2$ Remark: if $x^2-2 = \lceil A^n \rceil$ for all $n \in \mathbb{N}$, the "nearest square" condition can simply be checked by taking $A \to A^k$ for some large $k$ (I am too lazy but the only thing to be checked is that $\lceil A^n \rceil$ > 2).

$A$ is a biggest root of $x^2-(k^2-2)x+1=0$, where $k \ge 3 \in \mathbb{N}$

bora_olmez wrote: Take $A = \varphi^2$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. This choice of $A{}$ is motivated by the well-known theorem that a positive integer $N{}$ is a Fibonacci number if and only if at least one of $5N^2\pm 4$ is a perfect square.