Question: "However, how would you prove that a system of polynomial equations with fewer equations than variables has infinitely many solutions?"
Answer: There is a result in commutative algebra (look at the Matusumura book "Commutative ring theory", theorem 14.1, page 105) saying that if $(A,\mathfrak{m})$ is a noetherian local ring and $x_1,..,x_r$ is a system of parameters then
$$M1.\text{ }dim(A/(x_1,..,x_i))=r-i.$$
Intuitively view each element $x_i$ as an "equation" in $r$ variables, and $dim(A/(x_1,..,x_i))$ as the dimension of the "space of solutions" to the system of equations
$$M2. \text{ }x_1=\cdots =x_i=0,$$
then M1 says that the system M2 has an "infinite set of solutions" iff $i<r$.
"Intuitively" if $I:=(f_1, \cdots , f_l)\subseteq B:=k[x_1,..,x_n]$ is an ideal of polynomials in the variables $x_i$ over a field $k$, we let $A:=k[x_1,..,x_n]/I$ be the coordinate ring of the algebraic variety/scheme
defined by the set of polynomials $f_i$. The ring $A$ is a commutative unital ring reflecting properties of the "set of solutions" of the system of equations
$$M3.\text{ }f_1=\cdots =f_l=0.$$
We define $X:=Spec(A)$ as the "set of prime ideals" in the ring $A$ and give this set a topology (the Zariski topology). If the ideal $I$ is a prime ideal
it follows $X$ is an irreducible topolpogical space. The dimension of $X$
as an algebraic variety may be defined using the ring $A$: we define $dim(X):=dim(A)$ where $dim(A)$ is defined as the supremum of the set of strictly decreasing sequences of prime ideals
$$ \mathfrak{p}_r \subsetneq \mathfrak{p}_1 \subsetneq \cdots \subsetneq \mathfrak{p}_0 \subseteq A.$$
Hence with this definition, your system of equations should have a "finite set of solutions" iff $dim(A)=0$ - hence you must calculate the dimension $dim(A)$ in your case.
Given any maximal ideal $\mathfrak{m}\subseteq A$, it follows there is a maximal ideal $\mathfrak{n}\subseteq B$ mapping to $\mathfrak{m}$, and you may consider the localization $R:=B_{\mathfrak{n}}$ which is a local ring of dimension $n$. The image $g_i$ of the generators $f_i$ generates an ideal $J \subseteq R$ and if the ideal $J$ can be generated by a regular sequence
$J=(a_1,..,a_i)$ with $i<n$ it follows $dim(R/J)=dim(A) >0$. Hence the set
of solutions to the system is not finite. I believe there are algorithms implemented on computers that solves this problem, but I have no precise reference. In general it is difficult to check if an explicit ideal is a prime ideal, and it is also difficult to construct a regular sequence for an explicit ideal $J$.
Another more elementary approach is the Noether normalization lemma: It says that there is a set of element $t_1,..,t_d \in A$ with the property that the subring
$$S:=k[t_1,..,t_d] \subseteq A$$
is a polynomial ring, $dim(A)=d$ and $S \subseteq A$ is an integral extension.
Hence $dim(A)=0$ iff $k \subseteq A$ is an integral ring extension. Hence $dim(A)=0$ iff there is a finite set of elements $a_1,..,a_k\in A$ generating $A$ as $k$-algebra where $a_i$ satisfies a monic polynomial with coefficients in $k$. So your problem is reduced to constructing a set of integral elements $a_1,..,a_k$ generating $A$.
Lemma. The system M3 has a finite set of solutions iff $k \subseteq A$ is an integral ring extension iff $dim_k(A)< \infty$.
Proof. The first equivalence is proved above. The last equivalence follows from the fact that $A$ is finitely generated as $k$-algebra. Hence $k \subseteq A$ is an integral ring extension iff $dim_k(A)< \infty$. QED
Note 1. If you try to calculate $dim_k(A)$ "by hand" in some examples, you will find that it is a difficult problem to construct a basis for $A$ as $k$-vector space. I believe there are computer programs solving this problem.
To prove that $k \subseteq A$ is not an integral extension you must find an element $t\in A$ that generates a polynomial sub ring $k[t]\subseteq A$.
Note 2. If the base field $k$ is not algebraically closed, this method detects solutions in arbitrary finite extensions $k \subseteq K$. This is the "Hilbert Nullstellensatz". For a maximal ideal $\mathfrak{m}\subseteq A$ it follows the extension $k \subseteq A/\mathfrak{m}$ is a finite extension, and a maximal ideal $\mathfrak{m}\subseteq A$ corresponds to a solution to the system M3 in the field $\kappa(\mathfrak{m}):=A/\mathfrak{m}$.
Example 1. Let $k$ be the field of real numbers and let $f:=x^2+y^2+1\in k[x,y]$ and consider the "system"
$$S1.\text{ } f(x,y)=x^2+y^2+1=0.$$
The system S1 has no real solutions but it has many (a set of dimension 1) complex solutions. Given an arbitrary real number $\theta$ and let $x:=isin(\theta), y:=icos(\theta)$ it follows
$$x^2+y^2=(isin(\theta))^2+(icos(\theta))^2=-sin^2(\theta)-cos^2(\theta)=-1$$
hence $(isin(\theta), icos(\theta))\in \mathbb{C}^2$ is a solution to S1 for any $\theta$.
The field extension $\mathbb{R} \subseteq \mathbb{C}$ is finite and the dimension of the ring $A:=k[x,y]/(f)$ is 1, hence the system S1 has an infinite set of solutions (in the real and complex number field). For every $(isin(\theta), icos(\theta))$ you get a well defined surjective map
$$ \phi_{\theta}: A \rightarrow \mathbb{C}$$
defined by $\phi_{\theta}(x):=isin(\theta), \phi_{\theta}(y):=icos(\theta)$,
and the kernel $\ker(\phi_{\theta})\subseteq A$ will be a maximal ideal. Hence solutions to S1 correspond to maximal ideals in $A$.
Note 3. If you work with system M3 on a computer you should know that you are studying solutions to M3 in a finite field of large characteristic. This is because a computer has a finite memory. If your system M3 is defined by polynomials with rational coefficients $\frac{a_i}{b_i}$, it follows there is a large prime $p$ not dividing any of the $b_i$'s. The computer converts the system M3 to a system over $\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z}$ (or $\mathbb{F}_{p^r}$). Hence if you ask the question
$$\text{"Does M3 have a finite set of solutions?"}$$
to a computer, the computer will answer:
$$\text{It always has a finite set of solutions since we are working over a finite field $\mathbb{F}_p$}.$$
If $k$ is a finite field and you seek solutions to M3 in $k^n$ it follows the set of solutions $S$ to M3 is a subset $S \subseteq k^n$ of $k^n$ which is a finite set, hence $S$ is trivially a finite set. In a sense:
$$\text{A computer cannot "understand infinity".}$$
Example. If the memory $M$ of your computer $HAL$ has $2^n$ elements for $n \geq 1$ an integer, it cannot "work" with the ring of integers $\mathbb{Z}$ which has a countably infinite set of elements: There is no way to "embed" $\mathbb{Z}$ as a subset of $M$ since $M$ is a finite set. There is a number $m\geq 0$ with the property that an integer $a\in \mathbb{Z}$ with more than $m$ digits cannot be stored on $HAL$. Hence if $p,q$ are two prime numbers with more than $m$ digits it follows $p,q$ and the product $pq$
cannot be calculated by $HAL$. Hence $HAL$ cannot work with very large primes. There is an infinite set of prime numbers and only a finite set of prime numbers has less than $m$ digits. Hence there is an infinite set of prime numbers that cannot be calculated by $HAL$.
Hence by "understand" I mean the following: You may always construct two prime numbers $p,q$ and ask your computer $HAL$ to calculate $pq$, and your computer will not be able to do this because it does not have enough memory.
Question: "How would you characterize when a subset of equations "coincide" (you can combine several to get another, like linear dependence in linear systems)?"
Answer: If $I:=(f_1,..,f_l)$ is an ideal generated by polynomials $f_i$ and $J:=(f_{i_1},..,f_{i_n})$ is a subset of $I$ it follows (this is a definition) $I$ and $J$ "coincide" iff they generated the same ideal in $k[x_1,..,x_n]$.
