Hint: $(m+n)(m-n-1)=2018$, and the rest is easy.

$m^2=n^2 +m+n+2018 \Leftrightarrow (m-n)(m+n) - (m+n) = (m+n)(m-n-1) = 2018 = 2 \cdot 1009$. Note that $m+n > m-n-1$. So $m+n = 2018$ and $m-n-1 = 1$ or $m+n = 1009$ and $m-n-1= 2$. These cases give us, respectively, $(m,n) = (1010, 1008), (506, 503)$ (sorry for any latex mistake, i'm still learning it)