Divisibility problem involving the floor function I am trying to prove the following:

$$2^{m+1}\mid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$$ but $2^{m+2}\nmid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$ for every natural number $m$.

I tried to use induction on $m$ but am not sure how to use the induction hypothesis. (The base case is easy to verify). Is there some other way to approach this problem?
Thanks.

Consider the recurrence $x_0=2, x_1=20$ and $x_n=8x_{n-1}-4x_{n-2}$ for $n \geq 2$. It has characteristic polynomial $x^2-8x+4$, with roots $4 \pm 2\sqrt{3}=(1 \pm \sqrt{3})^2$, so $x_n=a(1+\sqrt{3})^{2n}+b(1-\sqrt{3})^{2n}$ for some contants $a, b$.
Using the values for $x_0, x_1$, we get: 
$$2=a+b$$
$$20=a(1+\sqrt{3})^2+b(1-\sqrt{3})^2=4(a+b)+2\sqrt{3}(a-b)=8+2\sqrt{3}(a-b)$$
$$a-b=\frac{12}{2\sqrt{3}}=2\sqrt{3}$$
$$a=\frac{2+2\sqrt{3}}{2}=1+\sqrt{3}, b=\frac{2-2\sqrt{3}}{2}=1-\sqrt{3}$$
Therefore $x_n=(1+\sqrt{3})^{2n+1}+(1-\sqrt{3})^{2n+1}$. Note that $x_n$ is an integer, and $-1<(1-\sqrt{3})^{2n+1}<0$, so $\lfloor(1+\sqrt{3})^{2n+1}\rfloor=\lfloor x_n-(1-\sqrt{3})^{2n+1} \rfloor=x_n$.
The problem is thus equivalent to proving that $2^{n+1} \| x_n$. We proceed by induction.
When $n=0$, clearly $2^1 \|2=x_0$. 
When $n=1$, clearly $2^2 \|20=x_1$.
Suppose that the statement holds for $0 \leq n \leq k, k \geq 1$. Then $2^{k} \|x_{k-1}$ and $2^{k+1}\|x_k$. Therefore $x_{k+1}=8x_k-4x_{k-1} \equiv 2^{k+2} \pmod{2^{k+3}}$, so $2^{k+2} \|x_{k+1}$.
We are thus done by induction.
