Solving $$2\cdot\binom{n}{5} = \binom{n}{4}+ \binom{n}{6}$$

gives $n=7, 14$.

Meanwhile, every term in the expanded form can be expressed as $$\binom{n}{k}x^{-\frac{4}{3}k}x^{n-k}=\binom{n}{k}x^{n-\frac{7}{3}k}$$

For a term to be constant, the following must be satisfied: $$n-\frac{7}{3}k=0 \implies k = \frac{3}{7}n$$

Setting $n=7$ and $14$, respectively, to the above relation finds two solutions: $\boxed{35}$ and $\boxed{3003}$.