You know that $x^{3}-7$ is irreducible over $\mathbf{Q}$ so the Galois group acts transitively on the set of roots, so $\mathbf{Q}(\sqrt[3]{7})$,$\mathbf{Q}(\zeta_{3}\sqrt[3]{7})$, $\mathbf{Q}(\zeta_{3}^{2}\sqrt[3]{7})$ are subfields of degree 3. And you know that these correspond to the subgroups of order 2 of $S_{3}$, and $S_{3}$ has exactly three such subgroups. Now $\mathbf{Q}(\zeta_{3})$ is a subfield of degree 2, and hence corresponds to a subgroup of order 3 in $S_{3}$, and $S_{3}$ has a unique such subgroup.
This isn't too hard to do, but the below method requires a little knowledge of the cubic resolvent $h$ of a quartic polynomial $g$, namely that $\operatorname{Gal}(h) \leq \operatorname{Gal}(g)$ and that the discriminants of $g$ and $h$ (essentially) coincide. (I can expand about this some if it would be helpful.)
If a polynomial $g$ is irreducible (which is a necessary condition for $\operatorname{Gal}(g) \cong S_{\deg g}$, and which we thus henceforth assume), then its Galois group acts transitively. The only transitive subgroups of $S_4$ are (up to conjugacy) $S_4, A_4, D_8, \Bbb Z_4, \Bbb Z_2 \times \Bbb Z_2$. We'll use the fact that the only groups among these whose order is divisible by $6$ are $S_4$ and $A_4$.
At least when the character of the underlying field $\Bbb F$ is not $3$, we may for simplicity of the below formulae make a linear change of variables so that $g$ has zero coefficient in its $x^3$ term and write (after dividing by the leading coefficient)
$$g(x) = x^4 + p x^2 + q x + r,$$
its resolvent cubic is
$$h(x) = x^3 - 2 p x^2 + (p^2 - 4 r) x + q^2,$$
and the discriminants of $g$ and $h$ coincide (perhaps up to an overall nonzero multiplicative constant), and are
$$D = 16 p^4 r − 4 p^3 q^2 − 128 p^2 r^2 + 144 p q^2 r − 27 q^4 + 256 r^3.$$
If $h$ is irreducible and its discriminant $D$ is not a square, then (1) $\operatorname{Gal}(h) \leq G := \operatorname{Gal}(g)$ is $S_3$, so $G$ has order divisible by $6$ and hence by the above is $S_4$ or $A_4$, and (2) since $D$ (the discriminant of $g$) is not a square, $G \not\leq A_4$ and hence $G \cong S_4$. For $g$ and $h$ to be irreducible, they must have nonzero constant terms and hence $q, r \neq 0$. For $p = 0$, the above formulae simplify to
$$h(x) = x^3 - 4 r x + q^2$$ and $$D = - 27 q^4 + 256 r^3,$$ so to find an example we can search for $q, r$ for which $h$ is irreducible and $D$ is nonsquare. For $\Bbb F = \Bbb Q$ the simple choice $q = r = 1$ satisfies these criteria (the first by the Rational Root Test), so, we have for example, that $$\operatorname{Gal}(x^4 + x + 1) \cong S_4 .$$
See these notes for more details (using the same notation). Also, note that these sorts of examples are generic in that, in a sense that can be made precise, most irreducible polynomials of degree $n$ in $\Bbb Q[x]$ have Galois group $S_n$.
Best Answer
Constructing ONE polynomial whose splitting field has a specific group as its Galois group is usual Inverse Galois Problem. That is possible for any cyclic group. And there exist infinitely many extensions with $C_8$ as Galois group: take any prime number $p$ of the form $8n+1$, then the $p$ cyclotomic extension has a degree 8 cyclic Galois extension. Generic Polynomial means ALLL these extension to be described through a ONE polynomial in many variables as the following example suggests:
Now look at the polynomial $x^2-y\in \mathbf{Q}[x,y]$ in two variables over rationals. Now any extension of Q with Galois group $C_2$ is obtained by giving special values for the variable y as a square-free integer. This two variable polynomial is called generic for $C_2$. What is claimed is there is no generic polynomial (multi-variable) where specialising values for the extra variables lead to EVERY possible one variable polynomial with $C_8$ as Galois group.