An integer cube will always leave a remainder of either $0, 1, 6$ when divided by $7$ (I suppose that I will leave this to you to prove), while $10000$ leaves a remainder of $4$ when divided by $7$.
So we see that if we divide $x^3 + 1000$ by $7$, it must leave a remainder of either $4, 5, 3$ (by summing up both of their remainders in the ring of integers modulo $7$).
On the other hand, $y^3$ is an integral cube. As disused earlier, all integer cubes must leave a remainder of either $0, 1, 6$ when divided by $7$.
Since the remainder of both sides when divided by $7$ will never be equal, there are no integer solutions.
A possible alternative would be to rearrange the given problem into:
$$y^3 - x^3 = 10000$$
Now, we factorize this to get:
$$(y-x)(y^2 + xy + x^2) = 10000$$
Over here, we can test all possible factors of $10000$ to see that no such integral $x, y$ exists.