Both sides of 2018

Notice that $2017 \equiv 1\pmod{4}$ and $2019^{2019} \equiv (-1)^{2019} \equiv -1 \equiv 3 \pmod{4}$ .

Hence our equation changes to $x^2+y^2 \equiv 3 \pmod{4}$ . But squares leave a remainder 0 or 1 when divided by 4. Thus $x^2+y^2 \equiv 0,1,2 \pmod{4}$ and $x^2+y^2 \not\equiv 3 \pmod{4}$ . So we conclude that the equation $2017x^2+y^2=2019^{2019}$ does not have integral solutions.

Notice that $2019 \equiv 3 \ (mod \ 4)$ . Then $2019^{2019} \equiv 2019^3 \ (mod \ 2017)$ Having said that, $y^2 \equiv 8 \ (mod \ 2017)$ but that would mean that $y \equiv \sqrt{8} \ (mod \ 2017)$ implying that there are not integer solutions to this equation.