Show that the ideal generated by two polynomials is actually equivalent to the ideal generated by a single polynomial

Well, there is no need to complete the last part of OP's attempt, but I am answering anyhow. The OP can try again to show that the greatest common divisor of $x^2+x+1$ and $x-3$ is $1$. That is, $$1=(x^2+x+1)\,p(x)+(x-3)\,q(x)$$ for some $p(x),q(x)\in\mathbb{C}[x]$. Thus, any $f(x)\in \mathbb{C}[x]$ is in the ideal generated by $x^2+x+1$ and $x-3$ as $$f(x)=f(x)\cdot 1=(x^2+x+1)\,p(x)\,f(x)+(x-3)\,q(x)\,f(x)\,.$$ But this is practically the same thing that mechanodroid did, and can be skipped entirely.

Hint:

Euclidean division gives

$$x^3-1 = (x^2-4x+3)(x+4) + 13(x-1)$$

so for $(x-1)q(x) \in \langle x-1\rangle$ we have$$(x-1)q(x) = \frac1{13}(x^3-1)q(x) - \frac1{13}(x^2-4x+3)(x+4)q(x) \in I$$