How is $x^2$ prime in this ring? Let $S$ denote the set of polynomials in $x$ with integer coefficients that have no linear term.  In other words, $S = \{ a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_0 \,|\, n \in \mathbb{Z}^{\geqslant 0}, a_i \in \mathbb{Z} \}.$

How is $x^2$ prime in this ring? It seems like $x^2 \mid x^6$ by $x^2 \cdot x^4 = x^6$. But $x^6 = x^3 \cdot x^3$ and $x^2$ does not divide $x^3$ in this ring. Help?
You're correct: $x^2$ is not a prime element of the ring $S$.  Indeed, this is a standard example of an element of a domain that is irreducible but not prime.
A frequent definition of primality in rings is the following:  a principal ideal $(w)$ is prime  $\iff$ $\forall u,v$ with $u.v \in (w) \implies u \in (w)$ or $v \in (w)$. Note that the the polynomials of the ideal $(x^2)$ have a zero constant term .Can you write two polynomials that $\notin (x^2)$  such that their product $\in (x^2)$?