As noted by Sheldon, the good starting point must be a critical point. There is indeed a theorem that says that if you have an attracting cycle, then at least one critical point must belong to its attraction basin.
Roughly, the idea of the proof is as follows : around an attracting fixed point, there is a linearizing coordinate $\phi$ such that $f(\phi(z))=\lambda \phi(z)$. That coordinate is defined only in a neighborhood of the attracting fixed point, however using the functional equation it satisfies it is possible to prolonge it until you meet a critical point.
Now if you have a cycle instead of a fixed point, just replace $f$ by $f^p$ to get the same result.
The Mandelbrot set is of mathematical interest because in complex dynamics, the global behaviour of the dynamics is generally ruled by the dynamics of the critical points. Thus knowing the dynamics of the critical points give you information on all of the dynamics. In the simplest family $z^2+c$, there is only 1 critical point (0) and so it is natural to look at what happens to its dynamics depending on $c$.
If you want to generalize the notion of Mandelbrot set for, say, cubic polynomials $z^3 + az+b$, you would have to look at the behaviour of two critical points, and so not only would you get a set in $\mathbb{C}^2$, you would also need to make a choice in your definition : are you looking at parameters where both critical points are attracted to a cycle, or one of them, or none ?
In your case, there is only one critical point, so your set is a reasonable analogue of the Mandelbrot set.
EDIT : note that the definition of the Mandelbrot set does not use attracting cycles, but depends on whether or not the critical point goes to infinity. It is conjectured (it's one of the most important conjecture in the field) that the interior of the Mandelbrot set is exactly composed of parameters for which the critical point is attracted to a cycle. However it is well known that in the boundary of the Mandelbrot set, you have no attracting cycles.
EDIT 2 : One of the most interesting features of the Mandelbrot set is that its boundary is exactly the locus of bifurcation, i.e. the set of parameters for which the behaviour of the dynamics changes drastically. If you choose any holomorphic family of holomorphic maps $f_\lambda$, you can also define the locus of bifurcation for this family. It has been proved that this set is either empty or contains copies of the Mandelbrot set.
Your plot looks great. I don't know any better way than what you are doing. As you zoom in deeper, it takes more and more iterations to get a reasonable picture. To some extent, you can use fixed point instead of floating-define maxint as $4$ and do fixed point multiplies and bit shifts to rescale, then add. You can also do your compares to $2$ in each direction instead of squaring. If the value is greater than $2$ in magnitude, it will be greater than $2$ in one axis soon.
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The Mandelbrot set is interesting because the location of a complex number $c$ in the Mandelbrot set yields information about the dynamics of the function $f_c$ and about the geometry of the Julia set $J_c$. This is due to the strong influence of the critical orbit (i.e. the orbit of the critical point zero). Generally, the orbits of of arbitrary points have no such overwhelming influence, so there's really no reason to think the sets $M_a$ to be particularly interesting.
Nonetheless, we might expect some superficial relationships. Here are some hasty claims - hasty in the sense that I've only given these a few minutes of thought.
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Note that this image suggests a gap on the real axis around $c=-2.5$. Indeed, if you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-2.5$, you'll find that the it diverges to $\infty$. If, however, you compute the orbit of $z=1$ under the iteration of $f_c(z)=z^2-3$, you'll find that it is just the bounded two cycle $1\to-2\to1$. Of course, the orbit of $1$ is fixed under $f_c(z)=z^2$. This suggests that the gap is genuine.