Romanian Mathematical Olympiad

Denote $m = \sqrt{n+3} + \sqrt{\sqrt{n+3}}$. Then $n + \sqrt{n+3} = (m - \sqrt{n+3})^2$, whence $(2m+1)\sqrt{n+3} = m^2 + 3$. Then there exists $p \in \mathbb{N}$ so that $n+3 = p^2$ and, since $p + \sqrt{n+p} \in \mathbb{N}$, there exists $q \in \mathbb{N}$ so that $n+p = q^2$.
Eliminating $n$ yields $p^2 - 3 = q^2 - p$, that is $4p^2 + 4p - 12 = 4q^2$, or $(2p+1)^2 - (2q)^2 = 13$, which is equivalent to $(2p+1-2q)(2p+1+2q) = 13$.
This gives $p = 3$, so $n = 6$.