We show that $x^2+y^2-1$ is irreducible in $\Bbb{Q}[x, y]$. It is the Exercise 9.4.11 in Dummit and Foote's Abstract Algebra. We view $x^2+y^2-1=y^2+(x^2-1)$ as a polynomial in $(\Bbb{Q}[x])[y]$.
Case 1. $y^2+(x^2-1)=[f(x)y+g(x)][h(x)y+i(x)]$. \begin{eqnarray*} &\text{If}& y^2+(x^2-1)\\ &=& [f(x)y+g(x)][h(x)y+i(x)] \\ &=& f(x)h(x)y^2+[g(x)h(x)+f(x)i(x)]y+g(x)i(x)\\ &\Rightarrow& f(x)h(x)=1\\ &\Rightarrow& f(x)=r\text{ and }h(x)=1/r\\ &\stackrel{g(x)h(x)+f(x)i(x)=0}{\Rightarrow}& (1/r)g(x)+ri(x)=0\\ &\Rightarrow& g(x)+r^2 i(x)=0\\ &\Rightarrow& g(x)=-r^2 i(x)\\ &\stackrel{g(x)i(x)=x^2-1}{\Rightarrow}& -r^2 i(x)^2=x^2-1\\ &\Rightarrow& r^2 i(x)^2=1-x^2\\ &\stackrel{2=\deg{1-x^2}=2\deg{ri(x)}}{\Rightarrow}& \deg{r i(x)}=1\\ &\stackrel{ri(x)=ax+b}{\Rightarrow}& (ax+b)^2=1-x^2\\ &\Rightarrow& a^2=-1, \text{ which is impossible because }a\in \Bbb{Q}. \end{eqnarray*}
Case 2. $y^2+(x^2-1)=f(x)[g(x)y^2+h(x)y+i(x)]$.
If $y^2+(x^2-1)=f(x)[g(x)y^2+h(x)y+i(x)]$, then $f(x)g(x)=1$ and $f(x)$ is a unit.
Note. Case 2 is necessary because we DON'T have the implication: If $f(x)$ can't be written as a product of two polynomials $g(x)$ and $h(x)$ whose degree less than $\deg{f(x)}$, then $f(x)$ is irreducible. For instance, $f(x)=2(x+1)\in \Bbb{Z}[x]$.