Why can two carbon atoms not form more than triple bond with each other?
Unfortunately, an answer to this requires a quick dip into orbital theory. You may have heard of s, p and d orbitals; carbon as a main group element of group 14 has the electronic configuration of $[\ce{He}]\,\mathrm{2s^2\,2p^2}$, meaning two of its electrons are in an s orbital and two in p orbitals in its atomic ground state. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. (This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.)

The bonding (or antibonding) orbitals that are generated by this process are labelled &#x3c3; or &#x3c0; orbitals. Note that these are the Greek letters corresponding to s and p: a &#x3c3; orbital is basically an s orbital streched in one direction and a &#x3c0; orbital is a p orbital streched in the same manner.

Due to a number of constraints that you will learn about soon enough™, a carbon–carbon triple bond $\ce{C#C}$ will consist of one &#x3c3; bond and two &#x3c0; bonds as shown in the image below.


Figure 1: $\ce{H-C#C-H}$ showing the &#x3c3; and &#x3c0; orbitals required to make the respective bonds. Image taken from jahschem.wikispaces.org.

See how two of the three possible p orbitals on each carbon (figure 1’s left half) overlap to form two &#x3c0; bonds; the third possible p orbital takes part in both the &#x3c3; bond to the other carbon and that to the hydrogen — it has been linear combined with the s orbital to form an sp hybrid orbital outside the scope of this answer.

The main point of interest here is that it is, for reasons of symmetry, not possible for more than two p orbitals of a single atom to form &#x3c0; bonds to a single other atom at any one time. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds.

The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding. Once you have a better understanding of orbitals, it may be worthwhile to look up quadruple bonds of transition metals such as in $\ce{[(H2O)Cr({\unicode{x3bc}-}OAc)4Cr(H2O)]}$ or dirhenium octachloride. The latter is particularly interesting because the fourth — &#x3b4;, in analogy to d — bond can be cleaved by irradiation causing a structural change that clearly shows there must have been a quadruple bond previously. To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though.

And finally, $\ce{C2}$ does exist; however, explaining it requires orbital theories on a much higher level than in the scope of this answer.