Let $I$ be an ideal of $R$, if $m(x)$ is a polynomial of $I$ with the smallest degree, how can we prove $(m(x))\subset I$?
For the additive part, it is easy. But for the multiplicative part, we need to pick $r$ from $R$ first. If $r$ is a constant, that will be clear. But can $r$ be polynomial? If so, how can we promise $(m(x))$ must be a subset of $I$?
I think that you refer $I$ as an ideal of $R[x]$ and not $R$. In general if $m(x)\in I$ then $(m(x))\subset I$ because $f(x)m(x) $ is a moltiplication of an element of the ideal $I$ with an element of $R[x]$ and so it belongs to $I$. The most interesting question is if it is possibile to get $(m(x))=I$. In general is not true for example in $\mathbb{Z}[x]$ the ideal $I=(x^2+1,2)$ is different by $(2)$ and so you have only $(2)\subset I$.
You can think that the problem in this case is that you can’t divide $x^2+1$ for $2$ because $\mathbb{Z}$ is not a Field and you can’t write his inverse. You can proof that if $R$ is a Field than $R[x]$ is a PID and so you have always that $(m(x))=I$.