Here's one way that I can think of. We want to show that $\omega^2\not\in \mathbb{Q}(20^{1/4}\omega)$. It follows that it's sufficient to show that $x^4+20$ is irreducible over $\mathbb{Q}(i)=\mathbb{Q}(\omega^2)$.
Any factorization in $\mathbb{Q}(i)$ will actually be a factorization in the Gaussian integers $\mathbb{Z}[i]$ by Gauss lemma. Assume that $x^4+20$ is not irreducible over $\mathbb{Z}[i]$. Since no root of $x^4+20$ is contained in $\mathbb{Z}[i]$ the polynomial needs to factorize as a product of quadratic polynomials. But the only quadratic polynomials dividing it in $\mathbb{C}[x]$ hence in $\mathbb{Z}[i][x]$ are $x^2\pm i\sqrt{20}$ which are not in $\mathbb{Z}[i][x]$, so it must be irreducible.
If $\omega^2=i\in \mathbb{Q}(20^{1/4}\omega)$, then it follows by the irreducibility of $x^4+20$ that $\mathbb{Q}(20^{1/4}\omega)/\mathbb{Q}(i)$ is of degree $4$, hence $\mathbb{Q}(20^{1/4}\omega)/\mathbb{Q}$ is of degree $8$, which is a contradiction.