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{align*} |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{align*}Alternative: We note that $ |z_1 - z_2| = |z_1| \cdot \left| 1 - \dfrac{z_2}{z_1} \right| $. Let $ u = \dfrac{z_2}{z_1} $, so that $ \dfrac1u + u = 1 $, or $ u^2 - u + 1 = 0 $. The solutions are $ u = \dfrac{1 \pm \sqrt{-3}}2 = \dfrac12 \pm i\dfrac{\sqrt{3}}{2} $. Then \begin{align*} |z_1 - z_2|^2 &= |z_1|^2 \cdot \left| 1 - \dfrac{z_2}{z_1} \right|^2 \\ &= 5^2 \cdot \left| -\dfrac12 \mp i\dfrac{\sqrt{3}}{2} \right|^2 \\ &= 25 \cdot 1, \end{align*}no matter which value of $ u $ we use. Therefore, $ |z_1 - z_2|^2 = \boxed{25} $.