Let $gcd(a,b) = t\Rightarrow (a,b)=(tr,ts) $ Therefore $t^3r^2s=2017t (r+s)$ $t^2r^2s=2017 (r+s) $ Note that $r^2s\mid 2017\implies r=1$ and $s\in\{1,2017\} $. $\textbf {Case 1.} $ $s=1$ This implies $a=b $ and from $$a^2=2017\cdot 2 $$Which indeed is a contradiction. $\textbf {Case 2.} $ $s=2017$ This implies $t^2=2017+1$ which means $2018$ is a perfect square, contradiction.