Let $a=x-z$ and $b=y-z$. Then we have $ab=3z+a+b-3$ and we need to prove that $3z^2+2z(a+b)+a^2+b^2 \geq 3$. Putting $z=(ab-a-b+3)/3$ in the inequality, it is left to show that $(ab-a-b+3)^2+2(a+b)(ab-a-b+3)+3(a^2+b^2) \geq 9$. The last one is equivalent to $(ab)^2-(a+b)^2+6ab+3(a^2+b^2) = 2(a+b)^2+(ab)^2 \geq 0$. Hence, we are done.

By AM--GM, $x^2+y^2+z^2=\frac{(x+y)^2}{2}+2+\frac{(x+y)^2}{2}+2z^2+z^2+1-2xy-2z^2-3$ $\geq 2(x+y)+2(x+y)z+2z-2xy-2z^2-3$ $=2\left(x+y+z-(x-z)(y-z)\right)-3=3.$

rogue wrote: Let $x,y,z$ be real numbers such that $(x-z)(y-z)=x+y+z-3.$ Prove that $x^2+y^2+z^2\ge3.$ Let $x=a+1$, $y=b+1$ and $z=c+1$. Hence, the condition gives $(a-c)(b-c)=a+b+c$ and $x^2+y^2+z^2\geq3\Leftrightarrow$ $\Leftrightarrow a^2+b^2+c^2+2(a+b+c)\geq0\Leftrightarrow a^2+b^2+c^2+2(a-c)(b-c)\geq0\Leftrightarrow$ $\Leftrightarrow2c^2+(a+b-c)^2\geq0$, which is obvious.