We must have $\sqrt{n+3}=m\Rightarrow n={{m}^{2}}-3\Rightarrow \sqrt{n+3}+\sqrt{n+\sqrt{n+3}}=m+\sqrt{{{m}^{2}}+m-3}\in \mathbb{N}$ hence we must have $\sqrt{{{m}^{2}}+m-3}\in \mathbb{N}$, but we have ${{m}^{2}}<{{m}^{2}}+m-3<{{(m+1)}^{2}},\forall m>3$ and for $m=3\Rightarrow n=6$