We have that $(a-1)(c-1)$ is prime, so one of $a$ and $c$ equals $2$. Since $ac$ is a perfect square, it follows that the other of $a$ and $c$ is also double a perfect square.

This is quite simple. Simply note that $b^2=ac$. Thus, \[ac-a-c+1=(a-1)(c-1)\]is a prime. This requires one of $a,c$ to be $2$. WLOG, let $a=2$. Note that then, $2\mid b$. Then, \[c=\frac{b^2}{2} = 2\left(\frac{b}{2}\right)^2\]Thus, $a,c$ are both twice a perfect square as claimed.