This is a consequence of the Noether normalization theorem, and some inductions. (It is not trivial.) First of all, there is a weak interpretation of the statement, which says all chains of primes in $A$ of maximal length have length $r$. The stronger interpretation is that any chain of primes in $A$ in which no further prime insertions are possible has length $r$. The stronger statement is true, but somewhat harder to prove. [The length of $P_0 \subsetneq \cdots \subsetneq P_r$ is declared to be $r$ here.]
The normalization theorem says there is a polynomial subring $B = k[y_1, \ldots , y_r]$ of $A$ so that $A$ is integral over $B$. Then the Going Up Theorem will tell you chains of primes in $B$ lift to chains in $A$, and chains of length $s$ in $A$ contract to chains of length $s$ in $B$. There are obviously chains of length $r$ in $B$, hence also in $A$. I'll prove by induction on $r$ that there are no chains of length greater than $r$ in any $A$ of trans deg $r$. If $r = 1$, this holds because $B = k[y]$ is a PID, and chains in $B$ and $A$ correspond. In general, the first non-zero prime in a maximal length chain in $B$ must be principal, generated by an irreducible polynomial $f$. But now it is easy that $B/(f)$ has transcendence degree $r-1$, so by induction chains going on up from $(f)$ in $B$ have length bounded by $r-1$.
This only proves the weak theorem. The stronger theorem uses the Going Down Theorem for integral extensions of an integrally closed domain. The polynomial ring $B$ is integrally closed, so Going Down applies. The point now is, if we have a chain in $A$ for which no insertions are possible, then by Going Down the smallest non-zero prime $P$ in that chain must intersect $B$ in a minimal prime $(f)$ in $B$. Now induction for the stronger statement applies to $B/(f) \subseteq A/P$.
I guess this isn't a 'self-contained' proof, so perhaps doesn't provide what you hoped for. But I don't believe there is any easier method.
May be you will find this method unnatural as well, but go to The Stacks Project, browse to the chapter "Exercises" (this is chapter $74$), and take a look at exercise $10.1$. You can also look at Spectrum of a linear operator on a vector space of countable dim in which I ask a question related to $10.1$. Note that this proves the Nullstellensatz only for $\mathbb{C}$, but has the advantage of using the language of linear algebra which you may prefer more/ find more intuitive.
The other option would be to convince yourself that Noether Normalization is saying something geometric (the Nullstellensatz is an easy consequence of Noether Normalization), and for this I can recommend Ravi Vakil's "Foundations of Algebraic Geometry" (found here) sections $11.2.3$ to $11.2.6$ in the March 23rd version of the notes, although this may be an overkill.
Ofcourse, @mbrown's comment of rephrasing the Nullstellensatz is perhaps the best way to think about it. Good luck!
Best Answer
In a commutative ring $A$, an ideal $I\subset A$ has two radicals:
a) The nilpotent radical $\sqrt I=Rad(I)=\cap_{I\subset P \:prime}$ which is the intersection of all prime ideals containing $I$.
b) The Jacobson radical $J(I)=\cap_{I\subset M \: maximal}$ which is the intersection of all maximal ideals containing $I$.
Now, there is a class of rings, deservedly called Jacobson rings, for which the two radicals coincide. Any algebra of finite type over a field is a Jacobson ring, and so there is no contradiction in the statements you mention in your question.
Warning Many rings are not Jacobson rings. For example a local domain (A,M) which is not a field is never Jacobson because the nilpotent ideal $(0)$ has $Rad((0))=(0)$ whereas $J((0))=M$.
So, beware: the local rings $\mathcal O_{X,x}$ of points $x\in X$ of the affine algebraic $k$-variety $X=Spec(A)$ are not Jacobson rings even though the finitely generated $k$-algebra $A$ is a Jacobson ring!