Why is $ (2+\sqrt{3})^n+(2-\sqrt{3})^n$ an integer?

Hint: Let $a=(2+\sqrt{3})^n$ and $b=(2-\sqrt{3})^n$. Then $ab=(4-3)^n=1$. Also $a^{-1}=b$. Now conclude that $a+b=a+a^{-1}$ is integral.

Consider the matrix

$$ A = \begin{bmatrix} 2 & 3\\ 1 & 2 \end{bmatrix} $$

Clearly $ A^n $ is a matrix with integer entries, therefore the trace $ \operatorname{tr} A^n $ is an integer. On the other hand, $ A $ is a diagonalizable matrix whose eigenvalues are $ \lambda_1 = 2 + \sqrt{3} $, $ \lambda_2 = 2 - \sqrt{3} $; therefore the eigenvalues of $ A^n $ are $ \lambda_1^n $ and $ \lambda_2^n $. Since the trace of a matrix is the sum of its eigenvalues, we conclude that $ \operatorname{tr} A^n = \lambda_1^n + \lambda_2^n $, and the quantity on the right hand side is an integer.

As an alternative to applying the binomial theorem (that is a fine way), the sequence given by $$ a_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n \tag{1}$$ fulfills $$ a_0=2,\qquad a_1=4,\qquad a_{n+2}=4a_{n+1}-a_n \tag{2} $$ hence $a_n\in\mathbb{Z}$ is trivial by induction.
In general, if $\eta,\xi$ are roots of a monic second-degree polynomial with integer coefficients, $$ \eta^{n+2}+\xi^{n+2} = (\eta+\xi)(\eta^{n+1}+\xi^{n+1})-(\eta\xi)(\eta^n+\xi^n) \tag{3}$$ proves just the same, since $(\eta+\xi)\in\mathbb{Z}$ and $\eta\xi\in\mathbb{Z}$ are consequences of Viète's formulas.