Substitute $x = a + 1$, $y = b + 1$, $z = c + 1$. Then \begin{align*} (a + 1)(b + 1)(c + 1) & = abc + 1,\\ (a + 1)(b + 1)(c + 1) & = (a - 1)(b - 1)(c - 1) + 2. \end{align*}In particular \[(a - 1)(b - 1)(c - 1) = abc - 1.\]These equations give \begin{align*} (ab + bc + ca) + (a + b + c) & = 0\\ -(ab + bc + ca) + (a + b + c) & = 0 \end{align*}so $a + b + c = ab + bc + ca = 0$. Thus $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0$ and so $a = b = c = 0$, implying $\boxed{x = y = z = 1}$. Remark. The substitution $(x, y, z) = (a+1, b+1, c+1)$ is motivated by noticing $x = y = z = 1$ is a solution.

Let $P(t)=(t-x)(t-y)(t-z)=t^3-at^2+bt-c$ so $P(1)=1-c$ and $P(2)=2-c$. Expanding, we get $a=b$ and $2a=b+3$, so $a=b=3$. Then $P(t)=(t-1)^3+1-c$ with all real roots. If $c\neq1$, then the roots of $P$ are $1+\sqrt[3]{c-1},1+\omega\sqrt[3]{c-1},1+\omega^2\sqrt[3]{c-1}$ where $\omega=e^{i\frac{2\pi}{3}}$, contradiction. So $c=1$ then $P(t)=(t-1)^3$ and thus $\boxed{x=y=z=1}$.

$-(xy+yz+xz)+(x+y+z)=0, -2(xy+yz+xz)+4(x+y+z)=6$ $x+y+z=3,xy+yz+xz=3$ $xy+yz+xz \leq \frac{(x+y+z)^2}{3} =3 \to x=y=z=1$

After we get that $x+y+z=3$ and $xy+yz+zx=3$ we can write $x=3-y-z$ and put it in second equation. We will now get this equation: $(3-y-z)(y+z)+zy=3$. Factorize this. $3y+3z-y^2-zy-zy-z^2+zy=3$ or simple $z^2+z(y-3)+y^2-3y+3=0$. We can solve this like quadratic equation for z. $z=\frac{3-y \pm \sqrt{(y-3)^2-4(y^2-3y+3)}}{2}$ $z=\frac{3-y \pm \sqrt{y^2-6y+9-4y^2+12y-12}}{2}$ $z=\frac{3-y \pm \sqrt{-3(y-1)^2}}{2} \implies -3(y-1)^2=0 \implies y=1$. We get that $z=1, x=1$. Finally, only solution is $(x, y, z) =(1, 1,1)$

A bit different solution. As above, we similarly get that $x+y+z=3$. Write $x$ as $3-z-y$ and obtain that $3y+3z-y^2-zy-zy-z^2+zy=3$. Rewrite this as $3y+3z-3-3yz=y^2+z^2-2zy\implies 3(y-1)(1-z)=(y-z)^2$. Similarly obtain that, $3(x-1)(1-y)=(x-y)^2$ and $3(z-1)(1-x)=(z-x)^2$. Let WLOG $x=1$, thus $y=z=1$. We confirm this as a solution to our equation system. Now suppose none of them are equal to $1$. Thus $a>1$ or $a<1$, where $a\in \{x,y,z\}$. By pigeon-hole principle, two of them are in same group, WLOG let $x$ and $y$ are in same group, hence $(x-1)(1-y)<0\leq (x-y)^2$. Here are no solutions.

My solution like @CantonMathGuy 's solution but the contradicition is different i just use x>0 y>0 z>0 fact and it is obvious. note that from $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9$ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq 3$ $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{xyz}$ and we find $xyz=1$ and it gives us $x=y=z=1$