What is GCD of $7^{3001}-1$ and $7^{3000}+1$?

Your remark that $7^{3001}-1=7\cdot (7^{3000}+1)-8$ is quite useful here. In fact, it implies that $$\gcd(7^{3001}-1,7^{3000}+1)=\gcd(8,7^{3000}+1).$$ Now note that $7^{3000}+1$ is even, but $$7^{3000}+1=(8-1)^{3000}+1\equiv (-1)^{3000}+1=2 \pmod{4}.$$

Can you take it from here?