jmc

Let $z=a+bi,$ where $a$ and $b$ are real numbers. Then $z^2=(a+bi{)}^2=a^2+2abi-b^2$ and $|z{|}^2=a^2+b^2,$ so $$a^2+2abi-b^2+a^2+b^2=3-5i.$$Equating real and imaginary parts, we get $$\begin{array}{rl}2a^2& =3,\\ 2ab& =-5.\end{array}$$From the first equation, $a^2=\frac{3}{2}.$ From the second equation, $$b=-\frac{5}{2a},$$so $$b^2=\frac{25}{4a^2}=\frac{25}{4\cdot 3/2}=\frac{25}{6}.$$Therefore, $$|z{|}^2=a^2+b^2=\frac{3}{2}+\frac{25}{6}=\boxed{\frac{17}{3}}.$$