Such huge powers gives me a headache!

$2^{200}-31 \cdot 2^{192}+2^{n}=2^{192}(2^{8}-31+2^{n-192})=2^{192}(225+2^{n-192})$ .

Since $2^{192}$ is a perfect square, $225+2^{n-192}$ must also be a perfect square.

This is equivalent to determining a Pythagorean triple where one term is $15$ and where the other term is a power of $2$ .

We find that the only Pythagorean triplet satisfying this condition is $(15,8,17)$ .

Therefore, $2^{n-192}=64=2^{6} \implies n=198$ is the only solution.

Suppose $2^{200} - 31 \cdot 2^{192 +} 2^{n} = m^{2}, m \in \mathbb{Z}$ . $\begin{aligned} 2^{200} - 31 \cdot2^{192} + 2^{n} & = m^2 \\ 2^{200} - (2^{5} - 1)2^{192} + 2^{n} & = m^{2} \\ 2^{200} - 2^{197} + 2^{192} + 2^{n} & = m^{2} \\ (2^{100})^{2} - 2 \cdot 2^{100} \cdot 2^{96} + (2^{96})^{2} + 2^{n} & = m^{2} \\ (2^{100} - 2^{96})^{2} + 2^{n} & = m^{2} \\ ((2^{4} - 1)2^{96})^{2} + 2^{n} & = m^{2} \\ (15 \cdot 2^{96})^{2} + 2^{n} & = m^{2} \\ m^{2} - (15 \cdot 2^{96})^{2} & = 2^{n} \\ (m - 15 \cdot 2^{96})(m + 15 \cdot 2^{96}) & = 2^{x} \cdot 2^{y} \end{aligned}$ Let $(m -15 \cdot 2^{96}) = 2^{x}$ and $(m + 15 \cdot 2^{96}) = 2^{y}$ and $x + y = n$ . Now, WLOG $x \leqslant y$ . $\begin{aligned}2^{y} - 2^{x} & = 30 \cdot 2^{96} \\ 2^{x}(2^{y - x} - 1) & =15 \cdot 2^{97}\end{aligned}$ Comparing, we get $\begin{aligned} 2^{x} & = 2^{97} \\ 2^{y - x} - 1 & = 2^{4} - 1 \end{aligned}$ $\Rightarrow x = 97, y = x + 4 = 101, n = x + y = \boxed{198}$