Non-isomorphicity of certain fields.

Question

I guess the following statement should be true, but I can not prove it, I can just prove the reverse implication. 

$\star$ Condition: Let $2\leq m$ be a natural number. Let's denote the multiplicative group of rational numbers by $\mathbb{Q}^{\ast^{}}$, and consider the subgroup of perfect $m^{\rm th}$-perfect powers $\mathbb{Q}^{\ast^{m}}$.

Let $t, t'$ be two elements in the quotient group $\dfrac{\mathbb{Q}^{\ast^{}}}{\mathbb{Q}^{\ast^{m}}}$ such that: the order of $t$ and $t'$ in the group $\dfrac{\mathbb{Q}^{\ast^{}}}{\mathbb{Q}^{\ast^{m}}}$ are equal to $m$, i.e. $m$ is the least positive integer such that $t^m \in \mathbb{Q}^{\ast^{m}}$.

(2): If The two sets $\{t, t^2, \cdots, t^{m}\}\neq\{t', t'^2, \cdots, t'^{m}\}$ are not the same in the group $\dfrac{\mathbb{Q}^{\ast^{}}}{\mathbb{Q}^{\ast^{m}}}$, Then

(3) the two extensions $\mathbb{Q}(\sqrt[m]{t})$ and $\mathbb{Q}(\sqrt[m]{t'})$ are non-isomorphic. 

We can reformulate it in the following way: Under the $\star$ Condition, [(2) implies (3)].

If $m=2$, or $m=3$, I can prove the statement. (For example see the answer by @ArturoMagidin to this question: Is $\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{3})$? )

Wild version of my Question: What are some ideas to show that two given fields are non-isomorphic?

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Irrelevant attempt:

I can just prove the reverse implication: Under the $\star$ Condition, [(3) implies (2)]. Instead of proving this, I will prove the contraposition, i.e. I will prove [$\neg(2)$ implies $\neg(3)$].

[$\neg(2)$ implies $\neg(3)$]: If the two sets $\{t, t^2, \cdots, t^{m}\}=\{t', t'^2, \cdots, t'^{m}\}$ are the same in the group $\dfrac{\mathbb{Q}^{\ast^{}}}{\mathbb{Q}^{\ast^{m}}}$, then the two fields extensions $\mathbb{Q}(\sqrt[m]{t})$ and $\mathbb{Q}(\sqrt[m]{t'})$ are isomorphic. 

Proof: If $\{t, t^2, \cdots, t^{m}\} = \{t', t'^2, \cdots, t'^{m}\}$, then there is an integer $1\leq k \leq m$ with $\gcd(k,m)=1$, such that $t^k= t' (\mod \mathbb{Q}^{\ast^{m}})$. Then clearly $\mathbb{Q}(\sqrt[m]{t}) = \mathbb{Q}(\sqrt[m]{t^k}) =\mathbb{Q}(\sqrt[m]{t'})$, and we are done.

Answer 1

$x^m-t$ is irreducible over $\Bbb{Q}$, when it stays irreducible over $\Bbb{Q}(\zeta_m)$ it works easily, for example when $\gcd(m,\phi(m))=1$.

If $\mathbb{Q}(t'^{1/m})\cong \mathbb{Q}(t^{1/m})$ then $\mathbb{Q}(t'^{1/m})= \mathbb{Q}(\beta)$ for some root $\beta$ of $t^{1/m}$'s minimal polynomial, ie. $\mathbb{Q}(t'^{1/m})= \mathbb{Q}(\zeta_m^r t^{1/m})$ for some $r$ and since it is real iff $r=0$ we get $\mathbb{Q}(t'^{1/m})=\mathbb{Q}(t^{1/m})$.

Next $t'^{1/m}=\sum_{l=0}^{m-1} a_l t^{l/m}\in \mathbb{Q}(t^{1/m})\cap \mathbb{Q}^{1/m}$ iff $\sum_{l=0}^{m-1} a_l \zeta_m^l t^{l/m}=\zeta_m^s\sum_{l=0}^{m-1} a_l t^{l/m}$ for some $s$ coprime with $m$, since the assumption at the beginning means that $t^{l/m},l\in 0\ldots m-1$ is a $\mathbb{Q}(\zeta_m)$-basis of $\mathbb{Q}(t^{1/m},\zeta_m)$, this implies that each $a_l \zeta_m^l - \zeta_m^s a_l=0$ ie. $t'^{1/m} =a_s t^{s/m}$.