Consider the ring of polynomials $\mathbb R[x,y]$ and $I$ the ideal $I=(1+x^2)$ generated by $x^2+1$. I want to determine whether $y^2+1+I$ is irreducible in $\mathbb R[x,y]/(I)$.
If $y^2+1+I$ were reducible we would find polynomials $h(x,y), g(x,y)$ and $s(x,y)$ with $\operatorname{degree}(h)\ge 1$ and $\operatorname{degree}(g)\ge 1$ such that $$y^2+1=h(x,y)g(x,y)+(1+x^2)s(x,y)$$ and this is impossible because the only factorization of $y^2+1=(y+i)(y-i)$ which happens in $\mathbb C[y]$ and not in $\mathbb R[y]$. So $y^2+1+I$ is irreducible in $\mathbb R[x,y]/(I)$.
Note that $\mathbb R[x]/(x^2+1)$ is isomorphic to the complex numbers by sending $x\mapsto i$. With that in mind, consider $(y-x)(y+x) + I$. What happens if you add $1$ and subtract $1$ to this expression?