What is this?

Let $m^2=2^{200}-2^{192}\times 31 +2^n=2^{192}\left(225+2^{n-192} \right)$ . Then $225+2^{n-192}=a^2$ for some positive integer $a$ . Let $k_1+k_2=k=n-192$ . Now $2^k=a^2-225=(a+15)(a-15)$ . We can express it in $\cases{2^{k_1}=a+15 \\ 2^{k_2}=a-15 }$ , where $k_1>k_2$ . The difference of the two equations yields $30=2^{k_1}-2^{k_2}$ , which implies that $2\times 15 = 2^{k_2}\left(2^{k_1-k_2}-1\right)$ . Clearly, $\cases{k_2=1\\2^{k_1-k_2}-1=15}$ and hence $\cases{k_2=1\\ k_1=5}$ . Finally, $k=6=n-192$ and thus $n=198$ .

I m not posting the whole solution just a hint:write 31 as 32-1 why??? I think you have got the answer actually this will make 2^ $5$ -1 therefor make our equation and thus by solving that we get $198$