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!
The second form is indeed, as you said, a more general statement. First, we prove that, if $\mathbb{K}$ is algebraically closed, every maximal ideal of $R = \mathbb{K}[x_1, x_2, \dots, x_n]$ is of the form $\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$. Obviously any such ideal is maximal. To prove the converse, consider a maximal ideal $\mathfrak{n}$ and the projection $\varphi: R\to R/\mathfrak{n}$. As you said, $R/\mathfrak{n} \simeq \mathbb{K}$ by the Nullstellensatz. Call $a_i$ the image of $x_i$. Then we easily see that $\mathfrak{m} = (x_1-a_1, \dots, x_n-a_n) \subset \ker(\varphi)$. By maximality of $\mathfrak{m}$, it must coincide with the kernel $\mathfrak{n}$.
We now pass on to $V(I)$. Note that, if $(a_1,\dots, a_n) \in V(I)$, by considering the evaluation morphism in $(a_1, \dots, a_n)$, we get that $I \subseteq M = (x_1 -a_1, \dots, x_n - a_n)$: indeed, $M$ is in the kernel, and by assumption $I$ is in the kernel, too. If, on the other hand, $I$ is proper, then it is contained in some maximal ideal and we also know, by above, that every maximal ideal of $R$ is in a one-to-one correspondence with $n$-tuples $(a_1, \dots, a_n)$ and that it vanishes when evaluated on such tuple. So, considering again the evaluation morphism in the $n$-tuple corresponding to one of the maximal ideals containing $I$, we get that $(a_1, \dots, a_n) \in V(I)$ since $I$ is contained in the kernel of such morphism. We have proven and can now formulate the following: $$V(I) \neq \varnothing \iff I \text{ is proper}$$
Your statement is the contrapositive, since $1 \in \sqrt{I} \implies 1 \in I$, which in turn that implies $I$ is not proper.
Best Answer
A lot of this just echoes what Alex Youcis has already said; I thought it would be good to have an answer.
What does it mean for a $B$-algebra $B \to A$ to be finitely generated (or "finite type")? The translation between this statement and "$A$ is a quotient of a polynomial ring $B[x_1, \dots, x_n]$ in finitely many variables by an ideal $I$" must become second nature. Of course, explicitly writing $A$ out in this way gives something slightly more: a choice of generators and a realization of $\operatorname{Spec} A$ inside $\mathbf{A}^n_B$ as a "closed subscheme". Now, $A$ is a field if and only if $I$ is maximal.
If $k$ is algebraically closed then you should be able to show that it has no nontrivial finite field extensions. Maybe it's good to turn things around and see where ideals like $\mathfrak{m} = (x_1 - a_1, \dots, x_n - a_n)$ turn up. What follows applies to any $k$ and describes the "$k$-rational points" of $\mathbf{A}^n_k$. The difference will be that there are closed points not of this form, e.g., the maximal ideal $(x^2 + 1, y - x)$ of $\mathbf{Q}[x, y]$.
Say I have a point $(a_1, \dots, a_n)$ in $k^n$. This induces a morphism of $k$-algebras $\xi\colon k[x_1, \dots, x_n] \to k$, $f \mapsto f(a_1, \dots, a_n)$ and the kernel is $\mathfrak{m}$. Work this out! Now show yourself that you can recover the point of $k^n$ from $\xi$.