$n^{2} + 96 = x^{2}.$
$x^{2} - n^{2} = 96.$
$(x-n)(x+n) = 96.$
Since, $x$ and $n$ both should be positive integer. $(x-n)$ and $(x+n)$ will be divisors of $96.$
By observation, $(x-n)$ should be smaller than $(x+n)$ because $x$ and $n$ are positive integers.
$(x-n) = k1 \rightarrow x= n+k1$
($x+n)= k2 \rightarrow n+k1+n = k2 \rightarrow 2n+k1 = k2 \rightarrow 2n = k2-k1 \rightarrow n=\left(\frac{k2-k1}{2}\right).$
As we have seen from above, $n = \left ( \frac{k2-k1}{2} \right )$
Therefore, for n to be a positive integer,$k2-k1$ should be even. That is, both should be odd or both should be even.
There are 6 pairs of divisors when multiplied becomes $96 = (1,96), (2,48), (3,32), (4,24), (6,16), (8,12) .$
Therfore, there are only 4 such possibilties. Phew!