Almost Perfect square

$2^{200}-(2^{5}-1)2^{192}+2^{n}$

= $2^{200}-2^{197}+2^{192}+2^{n}$

= $2^{192}(256-32+1+2^{n-192})$

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

Now, $225+2^{6} = 225 + 64 = 289= 17^{2}$

Hence, $n-192=6$

$\boxed{n=198}$

We note that:

$2^{200}-31\dot{}2^{192} = (2^8-31)\dot{}2^{192} = (256-31)\dot{}2^{192}$

$= 225\dot{} 2^{192} = 15^2\dot{}2^{192}$

Now consider the perfect square:

(15+x)^2\dot{} 2^{192} = (15^2+30x+x^2)\dot{} 2^{192} = 15^\dot{} 2^ {192} + (30x+x^2)\dot{} 2^{192}

$=2^{200}-31\dot{}2^{192} + (30x+x^2)\dot{} 2^{192}$

For $2^{200}-31\dot{}2^{192} + 2^n$ to be a perfect square, we equate

$(30x+x^2)\dot{} 2^{192} = 2^n$

and find that it is true when $x=2$ then we have:

$(60+4)\dot{}2^{192} = 64\dot{} 2^{192} = 2^6 \dot{} 2^ {192} = 2^ {198} = 2^n \quad \Rightarrow n = \boxed {198}$