jmc

From the equation $\frac{z_1}{z_2}+\frac{z_2}{z_1}=1,$ $$z_1^2+z_2^2=z_1{z}_2,$$so $z_1^2-z_1{z}_2+z_2^2=0.$ Then $(z_1+z_2)(z_1^2-z_1{z}_2+z_2^2)=0,$ which expands as $z_1^3+z_2^3=0.$ Hence, $z_1^3=-z_2^3.$

Taking the absolute value of both sides, we get $$|z_1^3|=|z_2^3|.$$Then $|z_1{|}^3=|z_2{|}^3,$ so $|z_2|=|z_1|=5.$ Then $z_1{\overline{z}}_1=|z_1{|}^2=25,$ so ${\overline{z}}_1=\frac{25}{z_1}.$ Similarly, ${\overline{z}}_2=\frac{25}{z_2}.$

Now, $$\begin{array}{rl}|z_1-z_2{|}^2& =(z_1-z_2)\overline{(z_1-z_2)}\\ & =(z_1-z_2)({\overline{z}}_1-{\overline{z}}_2)\\ & =(z_1-z_2)\left(\frac{25}{z_1}-\frac{25}{z_2}\right)\\ & =25+25-25\left(\frac{z_1}{z_2}+\frac{z_2}{z_1}\right)\\ & =25+25-25=\boxed{25}.\end{array}$$Alternative: We note that $|z_1-z_2|=|z_1|\cdot \left|1-\frac{z_2}{z_1}\right|.$

Let $u=\frac{z_2}{z_1}$, so that $\frac{1}{u}+u=1$, or $u^2-u+1=0$. The solutions are $u=\frac{1\pm \sqrt{-3}}{2}=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}.$ Then $$\begin{array}{rl}|z_1-z_2{|}^2& =|z_1{|}^2\cdot {\left|1-\frac{z_2}{z_1}\right|}^2\\ & =5^2\cdot {\left|-\frac{1}{2}\mp i\frac{\sqrt{3}}{2}\right|}^2\\ & =25\cdot 1,\end{array}$$no matter which value of $u$ we use. Therefore, $|z_1-z_2{|}^2=\boxed{25}.$