There goes a theorem which says that an ideal $I$ of $\mathbb{Z}[x]$ is maximal if and only if $I$ is of the form $\langle f(x),p\rangle$ where $p$ is a prime and $f(x)$ is irreducible modulo $p$.
Lets consider the ring $R=\mathbb{Z}[x]/\langle 2x+1,6\rangle$. In this ring we have relations $2x=-1$ so $3=3(-2x)=0$. Hence $3x=x-1=0$. Therefore $x=1$ where $f(x)$ is identified with the equivalent class. We get $R=\left\{0,1,2 \right\}$ so it seems that $R \simeq \mathbb{Z}_3$ which forces $\langle 2x+1,6\rangle$ be a maximal ideal in $\mathbb{Z}[x]$ which should be wrong.
My question is what the ring $R$ actually is and what am I missing.
Your correct calculations showed that $$ \langle 2x+1,6\rangle=\langle x-1,3\rangle $$ so it seems to me that the ideal is of the prescribed form.
The result you stated did not claim that the given form is the only way to list a set of generators for that ideal.
Anyway, $R\simeq\Bbb{Z}/3\Bbb{Z}.$