Modulo $10$: As LHS $\equiv1$, we get that $n$ is even. Modulo $4$: As RHS $\equiv1$, we get that $m$ is even. Hence $20^m$ and $19^n$ both are squares. We write $m=2p$. Since LHS is a square strictly smaller than $(20^p)^2$, we conclude that LHS is at most $(20^p-1)^2$. Hence $20^{2p} - 10(2p)^2 + 1 ~\le~ (20^p-1)^2 ~=~ 20^{2p}-2\cdot20^p+1$, which simplifies to $p^2\ge20^{p-1}$. (*) This last inequality (*) holds true only for $p=1$. This yields $m=2$ and $n=2$ as the only solution.

DylanN wrote: Determine all pairs $(m, n)$ of non-negative integers that satisfy the equation $$ 20^m - 10m^2 + 1 = 19^n. $$ $\color{blue}\boxed{\textbf{Answer: (2,2)}}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$20^m-10m^2+1=19^n...(\alpha)$$$$\Rightarrow (-1)^m-m^2+1\equiv 1\pmod{3}$$$$\Rightarrow m^2\equiv (-1)^m \pmod{3}$$If $m\equiv 1 \pmod{2}:$ $$\Rightarrow m^2\equiv 2 \pmod{3} (\Rightarrow \Leftarrow)$$If $m\equiv 0 \pmod{2}:$ In $(\alpha):$ $$\Rightarrow 0-0+1\equiv (-1)^n \pmod{5}$$$$\Rightarrow n \text{ is even}$$$$\Rightarrow 20^m \text{ and } 19^n \text{ are perfect squares}$$In $(\alpha):$ $$\Rightarrow 19^n<20^m$$$m=2t:$ $$\Rightarrow 19^n\le (20^t-1)^2$$$$\Rightarrow 20^m-10m^2+1\le 20^m-2\times 20^t +1$$$$\Rightarrow 20^t<5m^2$$$$\Rightarrow 20^t<20t^2$$$$\Rightarrow 20^{t-1}<t^2$$$$\Rightarrow t\le 1$$$$\Rightarrow m\in \{ 0,2 \}$$If $m=0:$ In $(\alpha):$ $$\Rightarrow 2=19^n (\Rightarrow \Leftarrow)$$If $m=2:$ In $(\alpha):$ $$\Rightarrow 400-40+1=19^n$$$$\Rightarrow n=2$$$$\Rightarrow (2,2) \text{ is the only solution}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$