 Solution

An integral domain is a nonzero commutative ring with identity and no zero divisors. A principal ideal domain is an integral domain in which every ideal is generated by one element.

The leading coefficient of a product of nonzero real polynomials is nonzero, so $\mathbb{R}[x]$ is an integral domain. The zero ideal is principal. In any nonzero ideal $J$, choose a nonzero polynomial $g$ of least degree. Divide an arbitrary $f\in J$ by $g$: $f=qg+r$ with $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Since $r\in J$, minimality forces $r=0$. Thus $J=(g)$ and $\mathbb{R}[x]$ is a principal ideal domain.

In $\mathbb{Z}[x]$ consider $J=(2,x)$. If $J=(h)$, then $h$ divides the constant polynomial 2, so $h$ is constant, and divisibility of $x$ forces $h=\pm 1$. That would give $J=\mathbb{Z}[x]$. But the constant coefficient of every element $2p+xq$ is even, so $1\notin J$. This contradiction proves $\mathbb{Z}[x]$ is not a principal ideal domain.