First, it can be shown using binomial expansion that $(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}$ is an even integer. This is because, in the expanded form, all the odd power of $\sqrt{3}$ are cancelled and the other terms doubled. $$\begin{align*} &1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}\\ =\ &\left(\binom{2n}{0} + \binom{2n}{1}\cdot(\sqrt{3}) + \binom{2n}{2}\cdot(\sqrt{3})^2 + \binom{2n}{3}\cdot(\sqrt{3})^3 +\cdots\right)\\ +\ &\left(\binom{2n}{0} - \binom{2n}{1}\cdot(\sqrt{3}) + \binom{2n}{2}\cdot(\sqrt{3})^2 - \binom{2n}{3}\cdot(\sqrt{3})^3 +\cdots\right)\\ =\ & 2\times\left(\binom{2n}{0} + \binom{2n}{2}\cdot(\sqrt{3})^2 + \cdots\right)\end{align*}$$

Therefore, $(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}$ can be assumed to be $2k$ where $k$ is a positive integer.

Meanwhile the fact that $0 < \sqrt{3}-1 < 1$ means $0 < (1-\sqrt{3})^{2n} < 1$.

Hence, the smallest integer that exceeds $(1+\sqrt{3})^{2n}$ must be $2k$. Now, $$\begin{align*} 2k=\ &(1+\sqrt{3})^{2n} + (1-\sqrt{3})^{2n}\\ =\ &(\sqrt{3}+1)^{2n} + (\sqrt{3}-1)^{2n}\\ =\ &((\sqrt{3}+1)^2)^n + ((1-\sqrt{3}-1)^2)^n\\ =\ &((4+2\sqrt{3})^n + (4-2\sqrt{3})^n)\\ =\ &(2^n(2+\sqrt{3})^n + 2^n(2-\sqrt{3})^n)\\ =\ &2^n((2+\sqrt{3})^n + (2-\sqrt{3})^n)\end{align*}$$

Similar to the reasoning above, the value of $(2+\sqrt{3})^n + (2-\sqrt{3})^n$ must be an even integer. Therefore, the $2^n((2+\sqrt{3})^n + (2-\sqrt{3})^n)$ must be a multiple of $2^{n+1}$.