Let $n$ be a positive integer and consider $f(x,n) = \frac{x^{(2n)}}{(2n)!}$ where $x^{(m)}$ is the rising factorial: $$\begin{align*} x^{(1)} &= x\\ x^{(2)} &= x(x+1)\\ x^{(3)} &= x(x+1)(x+2)\\ &\vdots \end{align*}$$ Now define $x_n$ by $$f(x_n,n) = -1$$ (what is its minimum polynomial)
We now can make a ring by extending the integers: $\mathbb Z[x_n]$.
For instance, $f(x_1,1) = -1$. This means solving $$-1 = x(x+1)/2$$ we get $$x = (-1 + \sqrt -7)/2$$ or $$ x = (-1 - \sqrt -7)/2$$
But we just write $x_1$ for both solutions.
Then our ring is $\mathbb Z[x_1]$ or simply said the ring $a + b x_1$ for integers $a,b$.
I believe $\mathbb Z[x_1]$ is a UFD (unique factoring domain).
In fact I conjecture:
For all $n>0$, $\mathbb Z[x_n]$ is a UFD.
Any (dis)proofs or counterexamples ?