[This answer has been edited to discuss the general case.]
I will assume that variety means irreducible (otherwise you could work on individual irreducible components). Then $f:X \to Y$ is dominant by assumption, and has closed image since its source is projective, thus it is surjective.
Thus $f^{-1}(Z) \to Z$ is also surjective.
Now, as described in e.g. this MO answer (or in a Hartshorne exercise, maybe in Section 4 of Chapter II), for the proper map $f$, the function $y \mapsto \dim f^{-1}(y)$ is upper semicontinuous, so if $z \in Z$, then the dimensions of $f^{-1}(z)$ are at least $\dim X - \dim Y$. This implies that $f^{-1}(Z)$ contains components of
codimension $1$. (The intuition is just that we can add the dimension of $Z$ and of a typical fibre. One way to make this precise is to note that if every component of $f^{-1}(Z)$ were of codimension at least $2$, then since it dominates $Z$, we would see that
a generic fibre would be of dimension $\dim X - \dim Y -1$, whereas we already noted that every fibre has dimension at least $\dim X - \dim Y$.)
In general you can't do better than this, because there are morphisms of $3$-folds (just to take an example) which are birational, but in which the preimage of some particular point $y \in Y$ is a curve. Then if you take $Z$ to be a generic codim'n one subvariety passing through $y$, its preimage will be the union of a codim'n subvariety of $X$ (the proper transform of $Z$) and the curve $f^{-1}(P)$. So in general you can't expect $f^{-1}(Z)$ to be equidimensional.
Note also that $f^{-1}(Z)$ can also contain multiple components of codimension $1$. (E.g. let $X \to Y$ be the blow up of a surface at a point, and let $Z$ be a curve that passes through the blown up point.)
Rereading the question, I see that the point of the question might be the equidimensionality, and so you might be interested in the counterexample involving $3$-folds. This MO question and answers gives one such example.
There is a corresponding statement for varieties, and it's exactly what you intuit.
To see why this is the case, recall we have an (inclusion reversing) correspondence between radical ideals and subvarieties of $V$. So asking if
$$\mathcal{V}(\mathfrak{a}) \supseteq \{ x \}$$
is asking if the variety of $\mathfrak{a}$ contains a point $x$ (that is, a minimal nonempty subvariety $\{ x \}$). By the inclusion reversing correspondence, this is the same as asking if
$$\mathfrak{a} \subseteq \mathcal{I}(\{x\})$$
that is, if $\mathfrak{a}$ is contained in the ideal associated to a point $x$ (that is, a maximal proper ideal).
Of course, we know that every proper ideal is contained in some maximal proper ideal (by Zorn's Lemma), and now by the nullstellensatz (since $k$ is algebraically closed) we know that maximal ideals of $k[V]$ are all of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$.
If you're not sure why we can apply the nullstellensatz here, it's because ideals in
$k[V] = k[x_1, \ldots, x_n] / I$ are exactly the ideals of $k[x_1, \ldots, x_n]$ containing $I$, so we get the claim by the "ordinary" nullstellensatz followed by the correspondence principle.
So, to put all the pieces together, how would you actually write up a proof of this fact? You should try it yourself (everything you need is in this answer) and check against the version I've written up under the fold:
Let $\mathfrak{a}$ be a proper ideal in $k[V]$. Then by Zorn's Lemma it is
contained in a maximal ideal $\mathfrak{m}$. By the nullstellensatz
(and the correspondence theorem) we know
$\mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$ for some
$(a_1, \ldots, a_n) \in k^n$.
Now taking $\mathcal{V}$ of the relation
$\mathfrak{a} \subseteq \mathfrak{m} = (x_1 - a_1, \ldots, x_n - a_n)$,
we find $\mathcal{V}(\mathfrak{a}) \supseteq \mathcal{V}(\mathfrak{m}) = \mathcal{V}((x_1 - a_1, \ldots, x_n - a_n)) = \{ (a_1, \ldots, a_n) \}$,
so the point $(a_1, \ldots, a_n) \in \mathcal{V}(\mathfrak{a})$. As desired.
I hope this helps ^_^
Best Answer
Question: "Are the irreducible components of a codimension one subvariety also codimension one?"
Answer: If $A:=k[x,y,z]$ and $I:=(x),J:=(y,z)$, it follows the ideal $IJ$ define a codimension one subscheme $X:=V(IJ)⊆\mathbb{A}^3$ (the dimension of the component of largest dimension is one). $X$ has two irreducible components: One of dimension $2$ and one of dimension $1$. Hence the answer is "no".