Let $I=\langle 3, x^3+x^2+1\rangle$ be an ideal in $\mathbb{Z}[x]$.
How do we see that $(x-1)\notin I$?
I have a roundabout way that is kind of clumsy.
Thanks a lot.
Suppose to the contrary $(x-1)\in I$.
Then $x-1=3g(x)+f(x)h(x)$, where $f(x)=x^3+x^2+1$.
Write $g(x)=f(x)q(x)+r(x)$, with $\deg r<3$.
Then $x-1=3(fq+r)+fh=f(3q+h)+3r$.
So $x-1-3r=f(3q+h)$. Since the LHS has degree strictly less than $3$, while $f$ has degree $3$, thus $x-1-3r=f(3q+h)=0$.
This implies $x-1=3r(x)$ which is a contradiction by considering the constants in the equation modulo $3$.
Your proof seems to be correct. Here is another way to prove the claim.
We know that $$\Bbb Z[x]/I \cong \Bbb F_3[X]/(X^3+X^2+1),$$ which has $3^3=27$ elements.
Assume that $x-1 \in I$. Then $(3,x-1) \subseteq I$, and since $(3,x-1)$ is a maximal ideal of $\Bbb Z[x]$, we would get $(3,x-1)=I$ or $I=(1)$. The latter case is not possible because $\Bbb Z[x]/I$ is not the trivial ring. Therefore we would have $$\Bbb Z[x]/I \cong \Bbb F_3[X]/(X-1) \cong \Bbb F_3$$ which has $3 \neq 27$ elements ; contradiction.