Factor? How?

Note that for all $x > 8$ , that $(x^2-1)^2=x^4-2x^2+1 < x^4-x^2+64 < x^4 = (x^2)^2$

Thus, in order for $x$ to be a solution, we must have $0 < x \le 8$ .

Checking all $x=1\to 8$ , we see that only $x=1,8$ result in $x^4-x^2+64$ . Thus our answer is $1+8=\boxed{9}$ .

Let $x^4-x^2+64=k^2$ for some $k\in\mathbb{N}$ .

$\rightarrow\displaystyle\,x^2=\frac{1+\sqrt{1-4\left(64-k^2\right)}}{2}$ (Ruling out the negative sign since $x^2>0$ )

Now since $x\in\mathbb{N}$ so discriminant must equal some perfect square, say $n^2$ for some $n\in\mathbb{N}$ . Then we have

$4k^2-255=n^2$

$\rightarrow\,\left(2k-n\right)\left(2k+n\right)=255$

Now since $2k-n<2k+n$ so we try out such factorizations of $255$ which satisfy this, thus giving us

$\left(2k-n,2k+n\right)=\left(1,255\right),\left(3,85\right),\left(5,51\right),\left(15,17\right)$

This yields $n=1,23,41,127\;\rightarrow\,x^2=1,12,21,64$ . Since $x\in\mathbb{N}$ so we rule out $12$ and $21$ thus giving $x^2=1,64\rightarrow\,x=1,8$ .

So the required sum is $1+8=\boxed{9}$ .