Question: So my question is: if we have a function f(x, y), how do we know if the curve f(x, y)=0 encloses a finite region or not. As an example, the curve $x^2+y^2-1=0$ encloses a finite region, while the curve $x^2-y^2-1=0$ doesn't.

Answer: 
Consider the following result from geometric analysis due to [Matthew Grayson (1987)](https://en.wikipedia.org/wiki/Mean_curvature_flow#Convergence_theorems):

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> Let $g:\mathbb{S}^1\to\mathbb{R}^2$ be a smooth embedding, then the mean curvature flow with initial data $g$ eventually consists exclusively of embeddings with strictly positive curvature.
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Mean curvature flow is the process of distorting an embedded smooth curve (or more generally embedded smooth manifold) at every point in the direction of its mean curvature vector. What this says is that if $g$ defines a curve defining a bounded region, then that curve will deform into a circle. Moreover this circle will shrink to a single point in finite time.
[![enter image description here](https://i.stack.imgur.com/m7vtQ.png)](https://i.stack.imgur.com/m7vtQ.png)

Image: [David Eppstein (CC0)](https://en.wikipedia.org/wiki/Curve-shortening_flow#/media/File:Convex_curve_shortening.png)

This if you have a function $f:\mathbb{R}^2\to\mathbb{R}^2$ whose level set is a disjoint union of smoothly embedded curves, you can independently perform mean curvature flow on each component of $f(x,y)=0$. If each component bounds a finite region, then in finite time this will produce a set of points. Thus the set of such functions $f$ is recursively enumerable.

(*Remark:* in fact it is the case that if two components of $f(x,y)=0$ are nested embeddings of $\mathbb{S}^1$, then performing mean curvature flow on both simultaneously, the two curves will not intersect at any point (until after the inner curve has shrunk to a point).

This doesn't completely answer your question, but hopefully it is somewhat useful.
First of all you can't call a region of $\mathbb{R}^2$ finite unless its cardinal is finite. What you are trying to say it is that the region is bounded (that is, that there is some $D(x\_0,r)$ such that the region is inside the disck)

Although this is not a very concrete question, I guess that the best I can offer you is the [Jordan Curve theorem](https://en.wikipedia.org/wiki/Jordan_curve_theorem). It states that if you have a jordan curve $C$ (a simple closed curve) in $\mathbb{R}^2$ then $\mathbb{R}^2\setminus C$ consists of exactly two connected components, one of the components is bounded (called the interior) and the other is unbounded (called the exterior). So every Jordan Curve 'Encloses a finite region of the space'.

In your examples $x^2+y^2-1=0$ is a circumference, which is a jordan curve and so the theorem holds. On the other hand $x^2-y^2-1=0$ is not a jordan curve because it is not a closed curve.