Question: My attempt: Since $R$ is not a field, there must be a nonzero element $a$ in $R$ that is not a unit. Let $I=(a,X)$, and we try to show that this ideal is not principal. Since $R$ is an integral domain, then $I=\left\{af(X)+cg(X): f(X),g(X)\in R[X]\right\}$. Suppose it has a generator $h(X)$, then $I=(h(X))$. Since $a\in I$ and $X\in I$, we have that $h(X)|a$ and $h(X)|X$.

How to proceed?

Answer: 
Typically, one then argues that $h(X)$ must be constant by degree considerations (since $R$ is a domain), and then argues a bit more to deduce the desired claim. However, I will suggest an easier route: note that the principal ideal $(X)$ is prime, since $R[X]/(X) \cong R$, but is not maximal, since $R$ is not a field. Since every prime ideal is maximal in a PID...
As $R$ is a domain, the degree of a product of polynomials is the sum of the factor degrees. Using $h(X)\mid a$, this tells us $\deg h\le \deg a=0$, hence $h$ is a constant. But then from $h(X)\mid X$, the leading coefficient of $h$ must divide the leading coefficient of $X$, i.e., the leading coefficient of $h$ - which is just the constant $h$ itself - is a unit.