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$\def\QQ{\mathbb{Q}}$This question seems unmotivated to me, but I might as well answer it: Yes, for any finite extension $K \supset \QQ$, there is some $\alpha \in K$ such that $\QQ(\alpha^n) = K$ for all nonzero integers $n$.

$\def\p{\mathfrak{p}}$Let $p$ be a prime which splits completely in $K$, say $p = \p_1 \p_2 \ldots \p_d$ with $d=[K:\QQ]$. We know that $p$ exists by CebatarovCebotarov. By the Chinese remainder theorem, there is some $\alpha$ such that the $d$ valuations $v_{\p_i}(\alpha)$ are all distinct. So, for any $n$, the $d$ valuations $v_{\p_i}(\alpha^n)$ are all distinct.

If $L$ is a proper subfield of $K$, then there is some pair $\p_i$ and $\p_j$ such that $\p_i \cap L = \p_j \cap L$. (Because $p$ can only break into at most $[L:\QQ]$ factors in $L$.) So, if $\alpha^n$ were in $L$, then $v_{\p_i}(\alpha^n)$ would equal $v_{\p_j}(\alpha^n)$. This contradiction shows that $\alpha^n$ is not contained in any subfield of $K$.

Remark: One can make the same argument work using the archimedean place, if it happens that $K$ is totally real. This is what is happening in the case of $\QQ(\sqrt{2})$.

1

$\def\QQ{\mathbb{Q}}$This question seems unmotivated to me, but I might as well answer it: Yes, for any finite extension $K \supset \QQ$, there is some $\alpha \in K$ such that $\QQ(\alpha^n) = K$ for all nonzero integers $n$.

$\def\p{\mathfrak{p}}$Let $p$ be a prime which splits completely in $K$, say $p = \p_1 \p_2 \ldots \p_d$ with $d=[K:\QQ]$. We know that $p$ exists by Cebatarov. By the Chinese remainder theorem, there is some $\alpha$ such that the $d$ valuations $v_{\p_i}(\alpha)$ are all distinct. So, for any $n$, the $d$ valuations $v_{\p_i}(\alpha^n)$ are all distinct.

If $L$ is a proper subfield of $K$, then there is some pair $\p_i$ and $\p_j$ such that $\p_i \cap L = \p_j \cap L$. (Because $p$ can only break into at most $[L:\QQ]$ factors in $L$.) So, if $\alpha^n$ were in $L$, then $v_{\p_i}(\alpha^n)$ would equal $v_{\p_j}(\alpha^n)$. This contradiction shows that $\alpha^n$ is not contained in any subfield of $K$.

Remark: One can make the same argument work using the archimedean place, if it happens that $K$ is totally real. This is what is happening in the case of $\QQ(\sqrt{2})$.