Use the fact that cubes are only congruent to -1, 0, 1 modulo 9 or 7

If I'm not mistaken, then $a^3 \equiv b^3 \pmod{167}$, so $a^{165} \equiv b^{165} \pmod{167}$. However, $a^{166} \equiv b^{166} \pmod{167}$ by FLT, so $a \equiv b \pmod{167}$, so $a \equiv b \pmod{167^3}$, a contradiction, as $2004$ is not divisible by $167^3$.

"However, $a^{166} \equiv b^{166} \pmod{167}$ by FLT" only works if both $a,b$ are divisible by $167$, or neither (but if one is, the other must be, so that is just a minor omission). But "so $a \equiv b \pmod{167}$, so $a \equiv b \pmod{167^3}$" is false. Why would it be? Only because $a^{3} \equiv b^{3} \pmod{167}$ is not enough reason. On the same line, you could have worked with $3$ rather than $167$ ... DAMP gave you a good hint; modulo $9$ the cubic residues are $0$ and $\pm 1$, while $2004 \equiv 6\pmod{9}$ (the hint modulo $7$ does not work, since $2004 \equiv 2\pmod{7}$).

Whoops! I'll start from the fact that $a \cong b (mod 167)$. Sorry for the minor error. I should have WLOGed that $a$ and $b$ aren't divisible by $167$, the single out that case afterwards. We know that $a \cong b (mod 167)$, so clearly since $a>b$, $a^3-b^3>2004$, contradiction. This is because $a^3-b^3=(a-b)(a^2+ab+b^2)$. $a-b \ge 167$, and clearly $a^2+ab+b^2>16$, so $a^3-b^3>2004$.

Case I: Obviously both a & b cannot be even as 2004 is not a multiple of 8. Case II: If one is odd and the other even. This is also not possible as 2004 is not odd. Case III: If both are odd, then (a-b ) is even and (a^2 + ab + b^2) is odd. So, (a-b ) =4 or 12 and (a^2 + ab + b ^2) = 501 or 167. Neither of them is possible. So, not possible.

Konigsberg wrote: Prove that the equation $a^3-b^3=2004$ does not have any solutions in positive integers. Just $\mod 9$, things.