jmc

Let $z=x+yi,$ where $x$ and $y$ are integers. Then $$\begin{array}{rl}z{\overline{z}}^3+\overline{z}{z}^3& =z\overline{z}(z^2+{\overline{z}}^2)\\ & =|z{|}^2((x+yi{)}^2+(x-yi{)}^2)\\ & =(x^2+y^2)(x^2+2xyi-y^2+x^2-2xyi-y^2)\\ & =(x^2+y^2)(2x^2-2y^2)=350,\end{array}$$so $(x^2+y^2)(x^2-y^2)=175.$

Since $x^2+y^2$ is positive, $x^2-y^2$ is also positive. So we seek the ways to write 175 as the product of two positive integers. Also, $x^2+y^2>x^2-y^2,$ which gives us the following ways: $$\begin{array}{cccc}{x}^2+y^2& x^2-y^2& x^2& y^2\\ 175& 1& 88& 87\\ 35& 5& 20& 15\\ 25& 7& 16& 9\end{array}$$The only possibility is $x^2=16$ and $y^2=9.$ Then $x=\pm 4$ and $y=\pm 3,$ so the four complex numbers $z$ are $4+3i,$ $4-3i,$ $-4+3i,$ and $-4-3i.$ When we plot these in the complex plane, we get a rectangle whose dimensions are 6 and 8.



The area of this rectangle is $6\cdot 8=\boxed{48}.$