Let $R$ be an integral domain and let $a,b \in R$. Let $aR$ and $bR$ denote the principal ideals of $R$ generated by $a$ and $b$ respectively.
If $a = bu$ then for every element $ar \in aR$ we have $ar = (bu)r = b(ur) \in bR$, so $aR \subseteq bR$. If $u$ is a unit then we can write $b = au^{-1}$ and so $bR \subseteq aR$ by a similar argument with $u^{-1}$ in place of $u$. Therefore $aR = bR$.
For the converse, assume that $aR = bR$. Then $a \in bR$, say $a = br_1$, and $b \in aR$, say $b = ar_2$. Then $a = br_1 = (ar_2)r_1 = a(r_2r_1)$. If $a = 0$ then $b = 0$ also, so $a = bu$ where $u$ is the unit $1$. Otherwise, by cancellation we have $1 = r_2r_1$, so $r_2$ and $r_1$ are units (because $R$ is commutative.) Therefore $a = bu$ where $u$ is the unit $r_1$.
For all $a,b \in \mathbb{Z}$ we have $a\mathbb{Z} = b\mathbb{Z}$ if and only if $a = \pm b$.
For all $f(x),g(x) \in \mathbb{Q}[x]$ we have $\langle f(x) \rangle = \langle g(x) \rangle$ if and only if $f(x) = c \cdot g(x)$ where $c \in \mathbb{Q}^*$ (meaning that $c$ is a nonzero constant polynomial.)
Consider the evaluation homomorphism $\phi_0 : \mathbb{Z}[x] \to \mathbb{Z}$ defined by $\varphi_0(f(x)) = f(0)$. Define the inverse image $N = \phi_0^{-1}[2\mathbb{Z}]$.
The ideal $N = \phi_0^{-1}[2\mathbb{Z}]$ is not a principal ideal of $\mathbb{Z}[x]$. Note that $N$ consists of all polynomials in $\mathbb{Z}[x]$ whose coefficients of $x^0$ are even. In particular, $2$ and $x$ are in $N$. Let $f(x) \in N$. We want to show that $N \ne \langle f(x) \rangle$.
If $f(x)$ is an integer (meaning a constant polynomial) then it is even, so the polynomials in $\langle f(x) \rangle$ only have even coefficients. Therefore $x \notin \langle f(x) \rangle$.
If $f(x)$ is not an integer, then $\deg f(x) \ge 1$. In this case, every nonzero element of $\langle f(x) \rangle$ also has degree $\ge 1$. In particular, $2 \notin \langle f(x) \rangle$.
Consider the principal ideal $\langle x \rangle$ of the ring $\mathbb{Z}[x]$. Note that this ideal consists of all polynomials in $\mathbb{Z}[x]$ whose coefficients of $x^0$ are zero.
This is a prime ideal: If $f(x)g(x) \in \langle x \rangle$ then the coefficient of $x^0$ in $f(x)g(x)$ is zero. The coefficient of $x^0$ in $f(x)g(x)$ is the product of the coefficient of $x^0$ in $f(x)$ with the coefficient of $x^0$ in $g(x)$. The coefficients are in $\mathbb{Z}$, which is an integral domain, so this means that the coefficient of $x^0$ in $f(x)$ is zero or the coefficient of $x^0$ in $g(x)$ is zero. Therefore $f(x) \in \langle x \rangle$ or $g(x) \in \langle x \rangle$.
This is not a maximal ideal: We have $\langle x \rangle \subsetneq N \subsetneq \mathbb{Z}[x]$ where $N$ is the ideal of $\mathbb{Z}[x]$ from problem 2.
(Another way to do this problem is to notice that $\langle x \rangle$ is the kernel of the evaluation homomorphism $\phi_0 : \mathbb{Z}[x] \to \mathbb{Z}$, which is surjective, so $\mathbb{Z}[x] / \langle x \rangle \cong \mathbb{Z}$ by the fundamental homomorphism theorem. Because $\mathbb{Z}$ is an integral domain, $\langle x \rangle$ is prime. Because $\mathbb{Z}$ is not a field, $\langle x \rangle$ is not maximal.)
Let $R$ be a principal ideal domain and let $P$ be a nontrivial prime ideal of $R$. We claim that $P$ is maximal. Let $N$ be an ideal of $R$ such that $P \subseteq N$. We want to show that $N = P$ or $N = R$.
By our hypothesis, $P$ and $N$ are both principal, say $P = \langle p \rangle$ and $N = \langle a \rangle$. Then $\langle p \rangle \subseteq \langle a \rangle$, so $p = ar$ for some $r \in R$. Because $P$ is nontrivial we have $p \ne 0$, so $a,r \ne 0$. Then $ar \in P$, so because $P$ is prime we have $a \in P$ or $r \in P$.
If $a \in P$, then $N = \langle a \rangle \subseteq P$, so $N = P$.
If $r \in P$, then $r = pr'$ for some $r' \in R$. Then $r = pr' = arr'$, so by cancellation (using the fact that $r \ne 0$) we have $1 = ar'$. Therefore $1 \in \langle a \rangle = N$, so $N = R$.
The natural guess for $N$ would be $\langle x^3 - 1 \rangle$, but this ideal is not maximal, so the factor ring $E = \mathbb{Q}[x]/N$ would not be a field. The problem is that the polynomial $x^3 - 1$ already has a zero in $\mathbb{Q}$, namely $1$. We can factor $x^3 - 1$ as $(x-1)(x^2 + x + 1)$.
Note that the polynomial $x^2 + x + 1$ has no zeroes in $\mathbb{Q}$. (One way to see this is to use the rational root theorem. Another is to prove that its evaluation at any rational number, or indeed at any real number, is positive.) Because $x^2 + x + 1$ has degree $2$, this means that it is irreducible. Therefore, if we define $N = \langle x^2 + x + 1\rangle$ and $E = \mathbb{Q}[x]/N$, then $N$ is a maximal ideal of $\mathbb{Q}[x]$ and $E$ is a field.
Define $\alpha = x + \langle x^2 + x + 1\rangle \in E$. As in the proof of Kronecker's theorem, $\alpha$ is a zero of $x^2 + x + 1$, which is a factor of $x^3 - 1$, so $\alpha$ is a zero of $x^3 - 1$. In other words, $\alpha$ is a cube root of unity. We have $\alpha \ne 1_E$ because $\alpha = x + \langle x^2 + x + 1\rangle$ and $1_E = 1 + \langle x^2 + x + 1\rangle$ and these are different cosets because $x-1 \notin \langle x^2 + x + 1\rangle$.
We are not quite finished, because there is a third cube root of unity in $E$. Indeed, if we define $\beta = \alpha^2$, then $\beta^3 = (\alpha^2)^3 = (\alpha^3)^2 = (1_E)^2 = 1_E$. Note that $\beta$ is not equal to $1_E$ or to $\alpha$: we have $\beta = x^2 + \langle x^2 + x + 1\rangle$, which is a different coset from $x + \langle x^2 + x + 1\rangle$ and $1 + \langle x^2 + x + 1\rangle$ because $x^2 - x, x^2 - 1 \notin \langle x^2 + x + 1\rangle$.