When $R$ is a field then surely we end up for no use for polynomial rings as $R[x]=R$? As you can prove that:
Suppose that $R$ is an integral domain. Then $(R[x])^{\times}=R^{\times}$
This should mean that if $R$ is a field i.e. everything is a unit but $0_R$ then every polynomial will have degree zero apart from the zero polynomial? Am I missing something?
Yes you are definitely missing something.
$\mathbb{R}[x]$ certainly isn't the same as $\mathbb{R}$. Say the polynomial $x^2 + 1$ would correspond to the real number $u$. What polynomial would then correspond to $u^{-1}$? There is no polynomial $q$ in $\mathbb{R}[x]$ with $q\cdot (x^2 + 1) = 1$...
The units in $\mathbb{R}[x]$ are the polynomials $(\neq 0)$ which only have a constant term. That's only a tiny part of the whole ring $\mathbb{R}[x]$.