Here's another way to think about the "line at infinity" and the "points at infinity"...
Think of the usual $XY$-plane as sitting inside of $3$-space, but instead of it sitting in its usual place, $\{(x,y,0) : x,y\in\mathbb{R}\}$, shift it up by $1$ so that it sits as the $z=1$ plane.
Now, you are sitting at the origin with a very powerful laser pointer. Whenever you want a point on the $XY$-plane, you shine your laser pointer at that point. So, if you want the point $(x,y)$, you are actually pointing your laser pointer at the point $(x,y,1)$; since you are sitting at the origin, the laser beam describes a (half)-line, joining $(0,0,0)$ to $(x,y,1)$.
Now, for example, look at the point $(x,0,1)$, and imagine $x$ getting larger. The angle your laser pointer makes with the $z=0$ plane gets smaller and smaller, until "as $x$ goes to infinity", your laser pointer is just pointing along the line $x$ axis (at the point $(1,0,0)$), and the same thing happens if you let $x$ go to $-\infty$. More generally, if you start pointing to points that are further and further away from the "origin" in your plane (away from $(0,0,1)$), the laser beam's angle with $Z=0$ gets smaller and smaller, until, "at the limit" as $||(x,y)||\to\infty$, you end up with the laser beam pointing along the $z=0$ plane in some direction. We can represent the direction with the slope of the line, so that we are pointing at $(1,m,0)$ for some $m$ (or perhaps to $(-1,-m,0)$, but that's the same direction), or perhaps to the point $(0,1,0)$. So we "add" these "points at infinity" (so called because we get them by letting the point we are shining the laser beam on "go to infinity"), one for each direction away from the "origin": $(1,m,0)$ for arbitrary $m$ for nonvertical lines, and $(0,1,0)$ corresponding to the direction of $x=0$, $y\to\pm\infty$.
So: the "usual", affine points, are the ones in the $z=1$ plane, and they correspond to laser beams coming from the origin; they are each of the form $(x,y,1)$ for some $x,y$ in $\mathbb{R}$. In addition, for each "direction" we want to include that limiting laser beam which does not intersect the plane $z=1$; those correspond to points $(1,m,0)$, or the point $(0,1,0)$ when you do it with the line $x=0$. So we get one point for every real $m$, $(1,m,0)$, and another for $(0,1,0)$. You are adding one point for every direction of lines through the origin; these points are the "points at infinity", and together they make the "line at infinity".
Now, put your elliptic curve/polynomial $F=Y^2 - X^3 - aX-b$, and draw the points that correspond to it on the $z=1$ plane; that's the "affine piece" of the curve. But do you also get any of those "points at infinity"?
Well, even though we are thinking of the points as being on the $XY$-plane, they "really" are in the $Z=1$ plane; so our equation actually has a "hidden" $Z$ that we lost sight of when we evaluated at $Z=1$. We use the homogenization $f = Y^2Z - X^3 - aXZ^2 - bZ^3$ to find it. Why that? Well, for any fixed point $(x,y,1)$ in our "$XY$-plane", the laser pointer points to all points of the form $(\alpha x,\alpha y,\alpha)$. If we were to shift up our copy of the plane from $Z=1$ to $Z=\alpha$, we'll want to scale everything so that it still corresponds to what I'm tracing from the origin; this requires that every monomial have the same total degree, which is why we put in factors of $Z$ to complete them to degree $3$, the smallest we can (making it bigger would give you the point $(0,0,0)$ as a solution, and we do need to stay away from that because we cannot point the laser pointer in our eye).
Once we do that, we find the "directions" that also correspond to our curve by setting $Z=0$ and solving, to find those points $(1,m,0)$ and $(0,1,0)$ that may also lie in our curve. But the only one that works is $(0,1,0)$, which is why the elliptic curve $F$ only has one "point at infinity".
Best Answer
I'm going to elaborate on the explanations given in the two comments. The homogeneous form of your equation, $Y^2Z−X^3−XZ^2−Z^3=0$ (letting $a=b=1$) becomes $z-x^3-xz^2-z^3=0$ if you dehomogenize at $Y=1$. Let's call the original $(x,y)$ plane the "object plane" and the new $(x,z)$ plane the "picture plane" (borrowing the language of perspective). The line at infinity is represented in the picture plane by the line $z=0$. The curve looks like this in the picture plane:
You can see the curve crossing (or actually, tangent to) the line at infinity at one point.
Now, how can we make this geometrically intuitive? As the other comment indicates, points on the line at infinity represent directions in the object plane; equivalently, families of parallel lines in the object plane all intersect at the same point at infinity. (This is really evident in classical perspective--look up one-point and two-point perspective.) In this case, all lines parallel to the y-axis meet at the point (0:1:0) at infinity.
They meet at the same point whether you go to infinity upwards or downwards. This is an aspect of the topology of the projective plane. Remember, any two distinct lines in the projective plane meet in one point, so two parallel lines meet in a single point at infinity.
Another approach: projective points of $\mathbb{R}P^2$ correspond to lines through the origin in $\mathbb{R}^3$. Any such line intersects the unit sphere in two antipodal points. This pair can represent the "projective point". We can throw away one hemisphere and represent projective points with single points on the hemisphere, except for points on the equator. There we have to identify antipodal points. Suppose the equator represents the line at infinity; so a line doesn't have two points on it "at infinity", but just one, and topologically every projective line is a circle (when the field is $\mathbb{R}$).
Finally, you might be wondering why the curve is tangent to the line at infinity. Imagine a perspective drawing of an infinite chessboard. All the north-south lines, $x=c$, come together at a single point at infinity. In the picture plane, this looks like a pencil of lines all radiating from the same point. Now, what about the tangent lines to your curve? The tangent at $(x_0,y_0)$ has larger and larger slope (in absolute value) as $y_0$ becomes larger and larger. But also, $x_0\rightarrow\infty$ as $y_0\rightarrow\infty$. So for $|y_0|$ large, the tangent line is close to one of these $x=c$ lines, one making a small angle with the line at infinity in the picture plane.
You'll find an even more discussion of this sort of thing at my Algebraic Geometry Jottings.