Regarding the Thurston norm and the ways that a three-manifold can fiber over the circle

The answer is: yes if the rank of $H_2(M;\mathbb{Z})$ is $\ge 2$; and no if the rank is $1$ because in that case there is up to isotopy a unique connected surface bundle structure on $M$. The proof uses nothing more than what is in Thurston's original article MR0823443, although there are probably multiple other ways to do it; I'll mention another proof due to Fried.

In the case of rank $\ge 2$, consider a fibered face, and let $C \subset H_2(M;\mathbb{Z})$ be the open cone over that face whose fiber has homology class in $C$. Fix one particular fibration over the circle. Since rank$\ge 2$, $C$ contains primitive elements of $H_2(M;\mathbb{Z})$ arbitrarily far from the origin, i.e. having arbitrarily large norm. So it remains to check that the norm evaluated on the homology class of a fiber equals minus the Euler characteristic of that fiber. To put it another way, you just need to check that a fiber of a fibration over the circle is norm minimizing in its homology class. This is a consequence of the theorem in Thurston's original article saying that any compact leaf of a taut transversely orientable foliation is norm minimizing; in more detail, what he proves is that the absolute value of the Euler class of the foliation evaluated on the compact leaf (which equals $|\chi(S)|$), equals the norm evaluated on that leaf.

Another proof I particularly like is Fried's formula for the restriction of the norm to $C$, from his article here. Consider the fiber $F$, consider its pseudo-Anosov monodromy $\phi : F \to F$, let $S_\phi \subset F$ be the set of singularities of $\phi$, and let $C_\phi \subset M$ be the suspension of $S_\phi$, so $C_\phi$ is a collection of oriented circles in $M$ where $M$ is regarded as the mapping torus of $\phi$. Turn $C_\phi$ into a 1-cycle by assigning to each of its components a coefficient $\frac{p}{2}-1$ where $p$ is the number of prongs of the corresponding singularity in $S_\phi$. The conclusion of Fried's theorem is that the norm is equal to the absolute value of the intersection number with $C_\phi$ (which is obviously equal to $|\chi(S)|$ by the Euler-Poincare formula for $\chi(S)$ expressed in terms of the singularities of $\phi$. Fried proves that intersection number with $C_\phi$ is equal to the norm throughout the entire open cone $C$.

See Autumn Kent's answer to this question.