In P. Samuel, *Anneaux factoriels*, pages 36-37, it's proved that $A=\mathbb R[X,Y]/(X^2+Y^2+1)$ is a UFD.

In the following we denote by $x,y$ the residue classes of $X,Y$ modulo $(X^2+Y^2+1)$. Thus $A=\mathbb R[x,y]$ with $x^2+y^2+1=0$.

*Lemma.* The prime ideals of $A$ are of the form $(ax+by+c)$ with $(a,b)\neq (0,0)$.

*Proof.* It's not difficult to see that these elements are prime: if $p=ax+by+c$ with $(a,b)\neq (0,0)$, then $A/pA\simeq\mathbb C$.

On the other hand, let $\mathfrak p$ be a non-zero prime ideal of $A$. Since $\dim A=1$ necessarily $\mathfrak p$ is maximal, so $A/\mathfrak p$ is a field. It's enough to show that $\mathfrak p$ contains an element of the form $ax+by+c$ with $(a,b)\neq (0,0)$. Note that $A/\mathfrak p=\mathbb R[\hat x,\hat y]$ and $\hat x,\hat y$ are algebraic over $\mathbb R$. Thus we have an algebraic field extension $\mathbb R\subset A/\mathfrak p$, and therefore $[A/\mathfrak p:\mathbb R]\le2$. In particular, the elements $\hat 1,\hat x,\hat y$ of $A/\mathfrak p$ are linearly dependent over $\mathbb R$.

Let $S\subset A$ be the multiplicative set generated by all prime elements $ax+by+c$ with $(a,b)\neq (0,0)$. The ring of fractions $S^{-1}A$ has no non-zero prime ideals, and therefore $S^{-1}A$ is a field, hence a UFD. Now we can apply Nagata's criterion for factoriality to conclude that $A$ is a UFD.

It's easily seen that $A$ also a PID (use this result.)

Let's prove that $A$ is not Euclidean.

*Lemma.* Let $A$ be a Euclidean domain. Then there is $p\in A-\{0\}$ prime such that $\pi(A^{\times})=(A/pA)^{\times}$, where $\pi:A\to A/pA$ is the canonical surjection.

*Proof.* If one considers $p$ a non-zero, non-invertible element with $\delta(p)$ minimal (here $\delta$ is an Euclidean algorithm), then $p$ is prime and for $\hat a\in A/pA$ invertible there is $u\in A$ invertible such that $u-a\in pA$ (write $a=px+u$ with $\delta(u)<\delta(p)$ and notice that $u$ is not $0$ - here one uses that $\hat a$ is invertible -, and necessarily invertible), that is, $\pi(A^{\times})=(A/pA)^{\times}$.

In our case $A^{\times}=\mathbb R^{\times}$. Since $A/pA\simeq\mathbb C$ for any prime $p\in A$, we have $(A/pA)^{\times}\simeq\mathbb C^{\times}$. If we assume that $A$ is Euclidean, then we get a surjective group homomorphism $\mathbb R^{\times}\to\mathbb C^{\times}$ which is also injective (see @zcn's comment below), a contradiction.