= Solution
Write $f$ as the <ring homomorphism> $A\to B$. By the permitted assumption, $B$ is a finitely generated $A$-algebra. We will prove that every element of $B$ is integral over $A$, including any nilpotent contribution. If $B=0$, there is nothing to prove.
We use the <integral closure as an intersection of valuation rings> in its general form: if an <integral domain> $R$ lies in a <field> $K$, then its <integral closure> inside $K$ is the intersection of all <valuation rings> of $K$ containing $R$. Here $K$ need not be the <fraction field> of $R$. One direction follows from <valuation rings are integrally closed>. For the other, if $0\ne x\in K$ is not integral, the <ideal> $x^{-1}R[x^{-1}]$ is proper: an expression for $1$ in that <ideal>, multiplied by a sufficiently large power of $x$, would be a monic equation for $x$. Choose a <maximal ideal> containing this <ideal>, and apply the <valuation domination lemma> to the corresponding <local ring> inside $K$. The resulting <valuation ring> $V$ contains $R$ and has $x^{-1}$ in its <maximal ideal>, so $x\notin V$. This proves the claimed intersection criterion. The <valuation domination lemma> is obtained by choosing, using <Zorn's lemma>, a maximal local overring dominating the given <local ring>; the usual alternative between adjoining $z$ and adjoining $z^{-1}$ forces this maximal overring to be a <valuation ring>.
The <Noetherian ring> $B$ has finitely many minimal <prime ideals> $\mathfrak q_1,\ldots,\mathfrak q_r$. Set
$$
D_i=B/\mathfrak q_i,\qquad K_i=\operatorname{Frac}(D_i),\qquad R_i=\operatorname{im}(A\to D_i).
$$
Take any <valuation ring> $V$ of $K_i$ containing $R_i$. The maps $B\to D_i\to K_i$ and $A\to R_i\to V$ form a square for the <valuative criterion for properness>. Its lift gives $B\to V$, whose composite with $V\hookrightarrow K_i$ is the original map. Consequently the image $D_i$ is contained in $V$. The intersection criterion proves that every element of $D_i$ is integral over $R_i$.
Fix $b\in B$. For each $i$, choose a monic polynomial $P_i(T)\in A[T]$ whose value at the image of $b$ in $D_i$ is zero; coefficients over $R_i$ can be lifted to $A$. The product $P(T)=\prod_iP_i(T)$ is monic and satisfies
$$
P(b)\in\bigcap_i\mathfrak q_i=\sqrt{(0)}.
$$
The <nilradical> of a <Noetherian ring> is nilpotent. Explicitly, choose finitely many generators $n_j$ with $n_j^{e_j}=0$; every product of $1+\sum_j(e_j-1)$ generators then contains a vanishing power. Thus $(\sqrt{(0)})^N=0$ for some $N$, and $P(T)^N$ is a monic polynomial annihilating $b$. Hence $B$ is integral over $A$.
Finally choose algebra generators $b_1,\ldots,b_s$, with monic equations of degrees $d_1,\ldots,d_s$. Reducing powers in these equations shows that the finitely many monomials
$$
b_1^{a_1}\cdots b_s^{a_s},\qquad 0\leq a_j<d_j,
$$
span $B$ as an $A$-<module>. Therefore $B$ is a finite $A$-<module>, proving
$$
\boxed{\text{a proper morphism between affine Noetherian schemes is finite}.}
$$
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