Proving that $2+\sqrt{2}$ is irreducible in $\mathbb{Z}[\sqrt{2}]$. I'm asked to show that $x=2+\sqrt{2}$ is irreducible in $\mathbb{Z}[\sqrt{2}]$ by using the norm map $$N:\mathbb{Z}[\sqrt{2}]\rightarrow \mathbb{Z}^+:a+\sqrt{2}b\mapsto |a^2-2b^2|$$
Now, if $x=yz$, then $2=N(x)=N(y)N(z)$ forcing wlog $N(y)=1$. I'm now stuck trying to show that $y$ must be a unit and would appreciate any help.

Use the definition of the norm. If $y=c+d\sqrt{2}$ then $N(y)=(c+d\sqrt{2})(c-d\sqrt{2})=1$. 
