The question can be rephrased as:
Is the triple $\ 1,\sqrt2,\sqrt4\ $ linearly independent over $\mathbb Q$?
And the answer is no (i.e., there is no such $v$).
Denote $\alpha:=\sqrt2$. Then $\alpha^3=2$. The main point is, that the polynomial $x^3-2$ (which defines $\alpha$ as its root) is irreducible over $\mathbb Q$: cannot be written as a proper product of polynomials of degree 1 and 2. (This can be shown directly..)
In other words, the field extension $\mathbb Q(\alpha)$ of $\mathbb Q$, is --as $\alpha$ is the root of the irreducible $x^3-2$-- per definition, is isomorphic to the quotient $K:=\mathbb Q[x]/(x^3-2)$ of polynomial ring $\mathbb Q[x]$. That is, the elements of $K$ are the polynomials, but $x^3-2 = 0$ is assumed (as the only rule) in $K$.
And, similarly as $\mathbb Q[x]$ has $1,x,x^2,x^3,x^3,\ldots$ as basis, $K$ has $1,x,x^2$ as a (standard) basis over $\mathbb Q$. ($x^3$ and above powers can be rephrased by $1,x,x^2$, using the rule: for example $x^3=2,\ x^4=2x,$ etc.)
The correspondence between $K$ and $\mathbb Q(\alpha)$ is simply given by $x\mapsto\alpha$.