jmc

Cubing the equation $z=x+yi,$ we get $$\begin{array}{rl}{z}^3& =(x+yi{)}^3\\ & =x^3+3x^{2}yi+3x{y}^2{i}^2+y^3{i}^3\\ & =x^3+3x^{2}yi-3x{y}^2-y^{3}i\\ & =(x^3-3x{y}^2)+(3x^{2}y-y^3)i.\end{array}$$Hence, $x^3-3x{y}^2=-74.$ We then have $$x(x^2-3y^2)=-74.$$Thus, $x$ must be a divisor of 74, which means $x$ must be 1, 2, 37, or 74. Checking these values, we find that the equation $x(x^2-3y^2)=-74$ has an integer solution in $y$ only when $x=1,$ and that integer solution is $y=5.$ Therefore, $z=\boxed{1+5i}.$