Let $F$ be a field with infinite elements such that for any integer $k$, there exists a $2^k$-th root of unity, denoted by $N$, with the property that $N$ is either an integer or can be represented as a vector of integers. More formally, we require that $N^{2^k} = 1$ and $N \in \mathbb{Z}$ or $N \in \mathbb{Z}^n$ for some $n \in \mathbb{N}$.
For example, if such a field $F$ exists, then for $k=8$, there is an element $N \in F$ such that $N^{256} = 1$, and $N$ can be represented with only integers. Complex numbers do not satisfy this requirement because, while there is an $N$ such that $N^{256} = 1$, that $N$ is in the form $(a + bi)$, where $a$ and $b$ are not integers.
One idea I had was to use Clifford Algebras, as I thought the extra dimensions might be helpful. However, I realized that there is no multivector $X$ such that $X^2 = i$ and $X$ has only integer elements. There is $\left(\frac{\sqrt{2}}{2}Z + \frac{\sqrt{2}}{2}XYZ\right)$, which, when squared, becomes $i$, but its elements are not integers.
Does such a field $F$ exist? If so, can you provide an example or a construction method? If not, can you provide a proof or an explanation for why it is not possible?