$[n \sqrt{2}] = [m (2+\sqrt{2})]$ for $m,n$ natural.

Here's a partial answer (let me know if partial answers are unwelcome on math.SE). The equation has no solutions in natural numbers $\mathbb{N}$. Suppose for contradiction that there exists $m,n\in\mathbb{N}$ with $[n\sqrt{2}]=[m(2+\sqrt{2})]$. Then there are three cases:

Case 1: $m=n$. Impossible because then $$[n\sqrt{2}]=[m(2+\sqrt{2})]=[n(2+\sqrt{2})]=[n\sqrt{2}]+2n$$ which implies that $2n=0$ and $n=0\notin\mathbb{N}$.

Case 2: $m>n$. Then if we write $m=n+l$ for some $l\in\mathbb{N}$, $$\begin{align*}[n\sqrt{2}]&=[(n+l)(2+\sqrt{2})] \\ & = [(n+l)\sqrt{2}]+2(n+l) \\ & >[n\sqrt{2}] \end{align*}$$ which is a contradiction.

Case 3: $m<n$. Then if $m=n-l$ for some $l\in\mathbb{N}$, we have $$\begin{align*}[n\sqrt{2}]&=[(n-l)(2+\sqrt{2})] \\ & = [(n-l)\sqrt{2}]+2(n-l). \end{align*}$$ But this won't get us anywhere (unless if someone can continue from here and prove me wrong).

And of course the result follows easily from the link Qiaochu provided: https://en.m.wikipedia.org/wiki/Beatty_sequence. In layman's terms, it states that $$[nr]\neq [m\frac{r}{r-1}]$$ for all positive integers $n,m$ and positive irrationals $r$. In particular, if $r=\sqrt{2}$, then $\frac{r}{r-1}=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2}$, so $$[n\sqrt{2}]\neq [m(2+\sqrt{2})]$$ for all $n,m$ positive integers.