After some calculations, we can easily know that $n_{4}\leq 6$. Now we only need to prove that $n_{k+1}\leq 2n_{k}+2$ for $k\geq 4$. Suppose the contrary, that $n_{k+1}\geq 2n_{k}+3$. Then $2^{n_{k}}\cdot \left(\phi-1\right)-\left[2^{n_{k}}\cdot \left(\phi-1\right)\right]\leq 2^{-n_{k}-2}$ Since $\phi=\frac{1+\sqrt{5}}{2}$, this means $2^{n_{k}-1}\cdot \sqrt{5}-\left[2^{n_{k}-1}\cdot \sqrt{5}\right]\leq 2^{-n_{k}-2}$ So $\left(2^{n_{k}-1}\cdot \sqrt{5}+\left[2^{n_{k}-1}\cdot \sqrt{5}\right]\right)\cdot \left(2^{n_{k}-1}\cdot \sqrt{5}-\left[2^{n_{k}-1}\cdot \sqrt{5}\right]\right)<1$. $5\cdot 2^{2n_{k}-2}-\left[\sqrt{5}\cdot 2^{n_{k}-1}\right]^2<1$, and we can easily see that this leads to a contradiction. So $n_{k+1}\leq 2n_{k}+2$ for $k\geq 4$, and we're done.