First observe that $(2+\sqrt{3})(2-\sqrt{3}) = 1$. If we let
$$x = (\sqrt{2+\sqrt{3}})^r = (2+\sqrt{3})^{\frac{r}{2}} \quad \text{we get}$$
$$\frac{1}{x} = (\sqrt{2-\sqrt{3}})^r = (2-\sqrt{3})^{\frac{r}{2}}.$$
Therefore, we have to find $r \in \mathbb{Q}$ such that $x + \frac{1}{x} = 14$, or equivalently, $x^2 - 14x + 1 = 0$. The two solutions to this equation are $7 \pm 4\sqrt{3}$. In the attempt to relate this to $2 \pm \sqrt{3}$ we quickly discover that $7 \pm 4\sqrt{3} = (2 \pm \sqrt{3})^2$. This means, $x$ has to be equal to
$$(2+\sqrt{3})^2 \quad \text{or} \quad (2-\sqrt{3})^2 = (2+\sqrt{3})^{-2}.$$
This implies $\frac{r}{2} = \pm 2$, i.e. $r = \pm 4$. A straightforward calculation shows that $r = 4$ and $r = -4$ are indeed solutions to the given equation.