Let $R$ be a commutative ring with $1$, and let $I ≤ R$ be an ideal. We call $\mathrm{Rad}(I) := \{r \in R: \exists n \in \mathbb{N}_0: r^n \in I\}$ the radical of $I$.
I now want to show that if $\mathrm{Rad}((0)) ≠ (0)$, then $R^\times \subsetneq R[X]^\times$ (where $R^\times$ and $R[X]^\times$ are the respective sets of invertible elements).
In other words, I want to find a polynomial $f \in R[X]$ with degree at least $1$ that is invertible (in $R[X]^\times$).
I first thought about choosing a polynomial like $f = r X + 1$, so that the factor $r X$ becomes $0$ when multiplied with sufficiently many $r$'s, i.e. we have something like $r^{n-1} f = r^{n-1}(r X + 1) = r^n X + r^{n-1} = r^{n-1}$. But then, the problem is that none of the powers of $r$ need to be invertible, so $r^{n-1}$, for all we know, might not be an element of $R^x$. Therefore, I expect it's not that simple.
I also thought that maybe, we can utilize the other properties of $Rad(I)$, i.e. that $Rad(I)$ is an ideal itself? But I'm not really sure how to continue at this point.
Let $R$ be a ring and $r\in R$ be a nilpotent element i.e $r^n=0$. Then: $$(1-r)(r^{n-1}+r^{n-2}+...+r+1)=1-r^n=1$$ Hence $1-r$ is invertible. Thus also $1-(-r)=1+r$ is invertible, just because $-r$ is nilpotent.
Your idea is not bad, but need slight modification. Pick $r\in R$ a nilpotent element. Then $rx\in R[x]$ is a nilpotent element. Hence: $$f(x)=1+rx$$ is invertible in $R$.