Nope. For example $n=1$ gives us $\sqrt{\lfloor n+\tfrac {2n}{3}\rfloor+7} = \sqrt{8}$.

Typo Maybe ? It shouldn't be that easy.

there are infintely many $n$s such that $\sqrt{\lfloor n+\tfrac {2n}{3}\rfloor+7}$ is not square. let $n=12k+9 \implies \sqrt{\lfloor n+\tfrac {2n}{3}\rfloor+7}=\sqrt{\lfloor \frac {24k+18}3 \rfloor +12k+9+7}=\sqrt{8k+6+12k+16}=\sqrt{20k+22} $ $20k+22\equiv 2 \pmod 4$ cant be a perfect square