$(*)$ Let $S/R$ be an extension of domains. Assume that for some $a\in R$, the ring $R[a]$ is Goldman. Then I want to show that $a$ is algebraic over $R$, whence $R$ is also a Goldman domain.
DEF A domain $R$ is said to be Goldman if there is $x\in R$ such that $K=R[x^{-1}]$ equals the fraction fields of $R$, equivalently $K$ is a finitely generated $R$ algebra, equivalently there is a nonzero element in the intersection of all nonzero primes in $R$.
There is a theorem that if $S/R$ is an algebraic extension of domains, with $S$ a finitely generated $R$-algebra then $R$ is Goldman iff $S$ is. Hence, it suffices in $(*)$ to show that $a$ is algebraic over $R$.
Actually you want to prove the following:
This is a well known result (see e.g. Kaplansky, Commutative Rings, Theorem 21). Just in case, I add a sketch of the proof: If $R[X]$ is a $G$-domain, then $K[X]$ is also a $G$-domain, where $K$ is the field of fractions of $R$. But $K[X]$ is in the same time a PID with infinitely many primes, a contradiction.