A quick glance through the sequence reveals $x_1=3,x_2=4,x_3=5,x_4=7,x_5=9$, and thus, $m=1,3$ works. We now show this is the only possibility. Let $m$ be such that $x_m=a,x_{m+1}=n$, and $x_{m+2}$ are in arithmetic progression. Then, $x_{m+2}=2n-a$. Now, $x_{m+1}=\lfloor \sqrt{2}x_n\rfloor$ yields that $n=\lfloor a\sqrt{2}\rfloor$, that is, $n<a\sqrt{2}<n+1$ (observe that all terms are integers, so the inequalities are strict). Moreover, $x_{m+2}=\lfloor \sqrt{2}x_{m+1}\rfloor$ implies that $2n-a<n\sqrt{2}<2n-a+1$. Now, combining these two, we get $2n-a<n\sqrt{2}<2a$, and thus, $2n<3a$. On the other hand, we also have $2n-a>n\sqrt{2}-1>(a\sqrt{2}-1)\sqrt{2}-1=2a-\sqrt{2}-1$, yielding $2n>3a-\sqrt{2}-1$. Thus, $2n\in\{3a-1,3a-2\}$. In the former, set $a=2k+1$, yielding $n=3k+1$. Using $n<a\sqrt{2}$, we get $3k+1<2\sqrt{2}k+\sqrt{2}$, that is, $k<\frac{\sqrt{2}-1}{3-2\sqrt{2}}$, yielding $a<6$. The latter can be handled analogously.