In general, I would like to prove that if $m>2$ is an integer, then $(5+\sqrt[m]2)^n$ is never an integer (unless for $n=0$).
First, I'm interested in the simple case $m=3$ (I already solved it for $m=2$). I actually think that $x_n=(1+\sqrt[3]2)^n$ is never a rational.
Here is what I tried.
Suppose that $r:=x_n \in \mathbb{Q}$ for some $n>0$. Then $\sqrt[n]{r} = 5+\sqrt[3]{2}$ has degree $3$ over $\Bbb Q$ and its minimal polynomial is $x^3-15 x^2+75 x-127$. It has to divide $X^n-r$ in $\Bbb Q[X]$. Therefore, we could find three integers $0≤r_1<r_2<r_3<n$ such that $$(X-\sqrt[n]{r}\zeta_n^{r_1}) (X-\sqrt[n]{r}\zeta_n^{r_2}) (X-\sqrt[n]{r}\zeta_n^{r_3}) = X^3-15 X^2+75 X-127$$ (and the product of the other linear factors $(X-\sqrt[n]{r}\zeta_n^{j})$ should have rational coefficients). I tried to compare the constant terms... without success. Even if Galois theory was useful for the case $m=2$, I'm not sure how to possibly use it here $(m≥3)$. Computing the norms $N_{\Bbb Q(\sqrt[m]{2})/\Bbb Q}$ could give necessary conditions on $n$ for $x_n$ being a rational number.
I also tried with field extensions and degrees, but I don't know anything about the irreducibility of $X^n-r$, for instance.
Thank you in advance!