Q is to prove that integer just above ($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers. Q is to prove that integer just above($\sqrt{3} + 1)^{2n}$ is divisible by $2^{n+1}$ for all n belongs to natural numbers.
In Q , by integer just above means that:
For an example , which is the integer just above 7.3 . It is 8. Then , Q wants you to prove that 8 is divisible by $2^{n+1}$.
\begin{equation}
\begin{array}{l}
(\sqrt{3}+1)^{2 n}=(4+2 \sqrt{3})^{n}=2^{n}(2+\sqrt{3})^{n}\\
=2^{n}(2+\sqrt{3})^{n}=2^{n}\left[^n C _{0}2^{n}+^n C_{1} 2^{n-1} \sqrt{3}+^n C_{{2}} 2^{n-2} \sqrt{3}^{2}+\right.\\
\begin{array}{l}
(\sqrt{3}-1)^{2 n}=(4-2 \sqrt{3})^{n}=2^{n}(2-\sqrt{3})^{n} \\
=2^{n}(2-\sqrt{3})^{n}=2^{n}\left[{ }^{n} c_{0} 2^{n}-n_{C}, 2^{n-1} \sqrt{3}+{ }^{n} c_{2} 2^{n-1} \sqrt{1}^{2} \ldots\right.
\end{array}\\
I+f+f)=2^{n}\left[2(\text { Integer) }]=2^{n+1}\right. \text { . Integer }\\
I+1=2^{n+1} \text { . Integer }
\end{array}
\end{equation}
In the image is the way this question is solved.
My Q from this method of solving is that if we notice at the end , we somehow got $ 2^{n+1}$. If the Q has taken some other value like $3^{n+3}$ or Something else. Then , it was not possible to prove this question.
What is another method to prove this Q or can you help me justify that the above method can be used for all kinds of Q.
Thank you.

Part of the problem is that the image was a bit obscured and so the transcription was a bit off. Here is a more accurate transcription with tags added for reference:

$$\require{cancel}
\left(\sqrt3+1\right)^{2n}=\left(4+2\sqrt3\right)^n=2^n\left(2+\sqrt3\right)^n\tag1
$$
$$
\!\!\!\!I+f=2^n\left(2+\sqrt3\right)^n=2^n\left[{}^nC_02^n+\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2+\dots\right]\tag2
$$
$$
\left(\sqrt3-1\right)^{2n}=\left(4-2\sqrt3\right)^n=2^n\left(2-\sqrt3\right)^n\tag3
$$
$$
f'=2^n\left(2-\sqrt3\right)^n=2^n\left[{}^nC_02^n-\cancel{{}^nC_12^{n-1}\sqrt3}+{}^nC_22^{n-2}\sqrt3^2-\dots\right]\tag4
$$

$$
I+\bbox[3px,border:1px solid black]{f+f'}=2^n[2(\text{Integer})]=2^{n+1}\cdot\text{Integer}\tag5
$$
$$
I+1=2^{n+1}\cdot\text{Integer}\quad\checkmark\tag6
$$

$(1)$ is just simple algebraic manipulation
$(2)$ is expanding via the Binomial Theorem ($I$ and $f$ are the integer and fraction parts)
$(3)$ is $(1)$ with the substitution $\sqrt3\mapsto-\sqrt3$; note that $\left(\sqrt3-1\right)^{2n}=\left(-\sqrt3+1\right)^{2n}$
$(4)$ is expanding via the Binomial Theorem
$(5)$ adding $(2)$ and $(4)$ cancels all of the terms with $\sqrt3$ to an odd power
$\phantom{\text{(5)}}$ and doubles all of the terms with $\sqrt3$ to an even power (these terms are integers)
$(6)$ since $0\lt\left(4-2\sqrt3\right)\lt1$, $f'\in(0,1)$ and by definition, $f\in[0,1)$
$\phantom{\text{(6)}}$ since $I+f+f'\in\mathbb{Z}$, we must have $f+f'=1$

Here is, in my experience, a more usual proof of this.
$\left(1\pm\sqrt3\right)^2=4\pm2\sqrt3$ are roots of $x^2-8x+4$. Therefore, the solution to the linear recurrence equation
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[3pt]
&=4(2u_{n-1}-u_{n-2})\tag7
\end{align}
$$
is
$$
\begin{align}
u_n
&=a\left(1+\sqrt3\right)^{2n}+b\left(1-\sqrt3\right)^{2n}\\[3pt]
&=a\left(4+2\sqrt3\right)^n+b\left(4-2\sqrt3\right)^n\tag8
\end{align}
$$
In particular, the sequence with $a=b=1$ starts out with
$$
u_0=2,u_1=8,u_2=56\tag9
$$
Note that $\left.2^{n+1}\,\middle|\,u_n\right.$ for $n=1$ and $n=2$. Then $(7)$ guarantees that
$$
\begin{align}
u_n
&=8u_{n-1}-4u_{n-2}\\[6pt]
&=8\cdot2^n\frac{u_{n-1}}{2^n}-4\cdot2^{n-1}\frac{u_{n-2}}{2^{n-1}}\\
&=2^{n+1}\left(4\frac{u_{n-1}}{2^n}-\frac{u_{n-2}}{2^{n-1}}\right)\tag{10}
\end{align}
$$
By induction, $(9)$ and $(10)$ show that $\left.2^{n+1}\,\middle|\,u_n\right.$ for all $n\ge1$.
Since $u_n\in\mathbb{Z}$,
$$
\begin{align}
u_n
&=\left(1+\sqrt3\right)^{2n}+\overbrace{\left(1-\sqrt3\right)^{2n}}^{\text{in }(0,1)}\\[3pt]
&=\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\tag{11}
\end{align}
$$
Therefore, $(10)$ and $(11)$ show that
$$
\left.2^{n+1}\,\middle|\,\left\lceil\left(1+\sqrt3\right)^{2n}\right\rceil\right.\tag{12}
$$
