It is possible to prove that there are finitely many solutions, as $\frac{\varphi(n)}{n} = \prod_{p \mid n} (\frac{p-1}{p}) < \prod_{p < n} (\frac{p-1}{p})= O(\frac{1}{log(n)})$, which means that for big enough $n$ we have $$\varphi(n)^{\frac{2}{3}}(\varphi(n)^{\frac{2}{3}}+1) >n^{\frac{4}{3}-\varepsilon}>2n$$As the bound we need for this problem is not that strong it should be possible to prove that the RHS is bigger than the LHS for all $n>k$ for a specific value of $k$ and check by hand small values of $n$.