Difference between a square and a power of 2

We need to look for where the modulo $7$ "cycles", (if any), for both the sequences $2^{n}$ and $n^{2}$ correspond.

For the $2^{n}$ sequence we note that $2^{n + 3} = 8*2^{n} \equiv 2^{n} \pmod{7}$ since $8 \equiv 1 \pmod{7}.$ So the modulo $7$ cycle here has length $3$ , and proceeds as $2,4,1,2,4,1,2,4,1,.....$

For the $n^{2}$ sequence, we first note that, since $(n + 7)^{2} \equiv n^{2} \pmod{7},$ the maximum length of any modulo $7$ cycle will be $7.$ Since the actual cycle proceeds as $1,4,2,2,4,1,0,$ we see that this cycle does indeed have length $7.$

The modulo $7$ cycle for $2^{n} - n^{2}$ will then have length $lcm(3,7) = 21,$ and proceeds as

$1,0,6,0,0,0,2,3,4,0,2,4,1,4,0,5,2,6,5,3,1.$

Thus for every successive sequence of $21$ integers starting at $n = 1$ there are a total of $6$ instances where $7|(2^{n} - n^{2}).$

Now since $10000 = 476 \times 21 + 4,$ and since of the first $4$ values in this last cycle there are $2$ instances where $7|(2^{n} - n^{2}),$ the answer to this question is $476 \times 6 + 2 = \boxed{2858}.$