\(\displaystyle x(1+r+r^2) = 741\)

Prime factors of \(\displaystyle 741\) are \(\displaystyle 3\), \(\displaystyle 13\), \(\displaystyle 19\), so \(\displaystyle x \in \{1, 3, 13, 19, 39, 57, 247, 741\}\). That's only eight cases to look at.

\(\displaystyle \begin{aligned} x=741 \implies r^2+r+1 &= 1 \implies r \in \{0,-1\} \implies z \in \{0, 741\} \\ x=247 \implies r^2+r+1 &= 3 \implies r \in \{1,-2\} \implies z \in \{247, 988\} \\ x=57 \implies r^2+r+1 &= 13 \implies r \in \{3,-4\} \implies z \in \{513,912\} \\ x=39 \implies r^2+r+1 &= 19 \qquad \text{has no integer solutions} \\ x=19 \implies r^2+r+1 &= 39 \qquad \text{has no integer solutions} \\ x=13 \implies r^2+r+1 &= 57 \implies r \in \{7,-8\} \implies z \in \{637, 832\} \\ x=3 \implies r^2+r+1 &= 247 \qquad \text{has no integer solutions} \\ x=1 \implies r^2+r+1 &= 741 \qquad \text{has no integer solutions} \end{aligned}\)

Thus we have \(\displaystyle z \in \{0,247,513,637,741,832,912,988\}\) and \(\displaystyle 0+247+513+637+741+832+912+988=4870\)

I'm guessing you guys missed the negative ratios.