I choose one of the two generator polynomials in binary Golay code with 23-length. For instance $g(x) = (x^{11} + x^{10} + x^6 + x^5 + x^4 + x^2 + 1)$ . Does there exist any $k \in N$ such that $g(x)$ divide $(x^5 + x + 1)^k$ over $\frac{Z_2(x)}{ \langle x^{23} -1 \rangle}$ ? If there is, find one.
P.S.-I hope that I've written the commands right and you can easily read what I wanted to. My apologies in case I haven't.
Unless I misunderstood the question the answer is negative for a rather simple reason.
Over $\Bbb{Z}_2$ we have the factorization $$ x^{23}-1=(x+1) \left(x^{11}+x^9+x^7+x^6+x^5+x+1\right) \left(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1\right). $$ This means that $x^5+x+1=(x^2+x+1)(x^3+x^2+1)$ has no common factors with $x^{23}-1$, and is therefore a unit in the quotient ring $R=\Bbb{Z}_2[x]/\langle x^{23}-1\rangle$. The units form a group, so we can immediately conclude that $(x^5+x+1)^k$ is a unit of $R$ for all $k\in\Bbb{N}$.
But $g(x)$ is a factor of $x^{23}-1$ so its (polynomial) multiples form a proper ideal $G$ of $R$ ($G$ is the Golay code, really). Obviously the ideal $G$ cannot contain any units of $R$ proving the claim.