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$2^{200} - 31.2^{192} + 2^{n}$

$= 2^{200} - ( 256 - 225)2^{192} + 2^{n}$

$= 2^{200} - 2^{8}.2^{192} + 225.2^{192} + 2^{n}$

$= 2^{192}( 225 + 2^{n - 192})$

for perfect square,

$225 + 2^{n - 192} = m^{2}$ ( for some integer m)

we can see lhs will be an odd number therefore rhs should also be a pefect square of odd number

$2^{ n - 192} = (m - 15)(m + 15)$

by the fundamental theorem of arithmetic , both should be power of 2 .

Since we have to find the minimum value therefore, the smallest perfect square of odd number after 225 is 289

$2^{n - 192} = 289 - 225$

$2^{n - 192} = 64 = 2^{6}$

$\boxed{n = 198}$