This is a classic, the standard proof is to consider
$ (\sqrt{3} + 1)^{2n} = x_n + y_n \sqrt{3}$
for integer $x_n$, $y_n$, (which you can see using Binomial theorem) and show that
$ (\sqrt{3} - 1)^{2n} = x_n - y_n \sqrt{3}$
again using Binomial theorem.
Now use the fact that $\sqrt{3} - 1 \lt 1$ and that
$(\sqrt{3} + 1)^{2n} + (\sqrt{3} - 1)^{2n} = 2x_n$
Thus the integer you are looking for is $2x_n$.
Get a recurrence for $x_n$ and $y_n$ and use induction.
To get the recurrence:
we have that $(\sqrt{3} + 1)^2 = 4 + 2\sqrt{3}$ and so
$ x_{n+1} + y_{n+1} = (x_n + y_n \sqrt{3})(4 + 2 \sqrt{3}) = (4x_n + 6y_n) + (2x_n + 4y_n) \sqrt{3}$
Thus
$ x_{n+1} = 4x_n + 6y_n$ $ y_{n+1} = 2x_n + 4y_n$
Write $x_{n+2} = 4x_{n+1} + 6y_{n+1}$ and eliminate $y_n$ and $y_{n+1}$
(I believe it comes to $x_{n+2} = 8x_{n+1} - 4 x_n$)