We therefore either have y = 1 or x = z. In the first case our system reduces to x + z = 3 and xz = 2. Solving the first equation for z and substituting into the second yields the quadratic equation x2 - 3x + 2 = 0 which gives (x,y,z) = (1,1,2) or (2,1,1).
In the second case our system becomes xy + x = 3 and x2 + y = 3. Solving the first equation for y, substituting into the second and expanding gives us the cubic equation x3 - 4x + 3 = 0. Fortunately, the left-hand side factors to give (x - 1)(x2 + x - 3) = 0. This gives us (x,y,z) = (1,2,1) and x = y = z = (-1 ± Sqrt[13])/2.