Its square root is an integer

We require that $n^{2} + 96 = m^{2}$ for some integer $m.$ This can be rewritten as

$m^{2} - n^{2} = 96 \Longrightarrow (m - n)(m + n) = 96.$

Now let $a = m - n, b = m + n \Longrightarrow a + b = 2m.$ So we are looking for factor pairs of $96$ whose sums are even. Also, since we square $m$ in the original equation we can assume that $a,b$ are positive, (implying that $m$ is positive), without fear of missing any potential solutions for $n.$ With that said, we then have that $a \lt b,$ giving us the solution pairs

$(a,b) = (2,48), (4,24), (6,16), (8,12),$

which in turn correspond to the pairs

$(m,n) = (25,23), (14,10), (11,5), (10,2).$

Thus there are $\boxed{4}$ positive integral values of $n,$ namely $2,5,10$ and $23.$

I thought of $n^2+96$ as the square of a sum: $(n+m)^2=n^2+2nm+m^2$ .

Therefore, $96=2nm+m^2$ , and we find all the different ways of writing $96$ as the sum of a square plus an even number.

There are four ways of doing that:

$32+8^2 \implies 2\cdot 8 \cdot n=32 \implies \boxed{n=2}$

$60+6^2 \implies 2\cdot 6 \cdot n=60 \implies \boxed{n=5}$

$80+4^2 \implies 2\cdot 4 \cdot n=80 \implies \boxed{n=10}$

$92+2^2 \implies 2\cdot 2 \cdot n=92 \implies \boxed{n=23}$