I would like some elaboration for the bound appearing in this answer.
Namely, let $u\in \mathbb Z[\sqrt 2]^\times $ be an invertible element of the ring $\mathbb Z[\sqrt 2]$ with $u>1$. Then, there exists some nonnegative integer $k\in \mathbb Z$ such that $$(1+\sqrt{2})^k\le u <(1+\sqrt{2})^{k+1}.$$ Do we use any Calculus argument here? Could you please give me a hand please?
Let $r = 1 + \sqrt{2}$ and let $u \in [1,\infty)$ be arbitrary. Since $r > 2$, we have $r^k \geq 2^k > k$ for all positive integers $k$, so the increasing sequence $r^0, r^1, \dots$ is unbounded. Thus, the set $\{n \in \mathbb{N} : r^n \leq u\}$ is bounded (i.e. finite) and nonempty (because $r^0 = 1 \leq u$). Let $N$ be the maximum element of this set. By maximality, $r^{N+1} > u$, so we have
$$(1+\sqrt{2})^N \leq u < (1+\sqrt{2})^{N+1},$$
as desired.