Is there a third dimension of numbers? Is there a third dimension of numbers like real numbers, imaginary numbers, [blank] numbers?
 A: Alas, there are no algebraically coherent "triplexes". The next step in the construction as has been said already are "quaternions" with 4 dimensions.
Many young aspiring mathematicians have tried to find them since Hamilton in the 19th century. This impossibility links geometric dimensionality, fundamental properties of polynomial equations, algebraic systems and many other aspects of mathematics. It is really worth studying.
A quite recent book by modern mathematicians which details all this for advanced college undergraduates is Numbers by Ebbinghaus, Hermes, Hirzebruch, Koecher, Mainzer, Neukirch, Prestel, Remmert, and Ewing.
However, the set of quaternions with zero real part is an interesting system of dimension 3 with very interesting properties, linked to the composition of rotations in space.
A: You might look up quaternions.
A: In addition to complex numbers and quaternions, you might want to look up Clifford Algebras which encapsulate both and extend to arbitrary dimension.  Complex and quaternions are sub-algebras of the Clifford Algebras over $\mathbb{R}^2$ and $\mathbb{R}^3$ respectively.
A: Every finite-dimensional division algebra over $\mathbb{R}$ is one of $\mathbb{R}$, $\mathbb{C}$ or $\mathbb{H}$. This is what is called the Frobenius Theorem. You may refer to here for details.
A: It is a nice consequence of the hairy ball theorem that certain 3-dimensional number-analogues cannot exist.
Suppose we can define nice addition and multiplication on $V\cong \Bbb R^3$. 
Under some natural conditions on these operations (see below), one finds that there is an element $i\in V$ (named accordingly to the complex unit $i\in\Bbb C$) so that multiplication with $i$ in $V$ behaves like a $90^\circ$ rotation.
That is, $x$ and $ix$ are orthogonal to each other for all $x\in V$.
But then $\mathbb S^2 \to \mathbb S^2, x\mapsto ix$ defines a unit vector field on the 2-sphere, which is known to not exist by the hairy ball theorem.
The same argument works for all odd-dimensional cases, but is not strong enough to exclude most even-dimensional cases (as is necessary by the other answers).

Some details
Some natural requirements on such a number system are the following (there are others, but many turn out to be equivalent or stronger):


*

*The addition is component-wise, that is, $(x,y,z)+(x',y',z')=(x+x',y+y',z+z')$.

*The number system extends the real numbers in a natural way, e.g. each real numbers $x\in\Bbb R$ is represented  by $(x,0,0)\in V$ or the like (as $x\mapsto x+0i$ embedds $\Bbb R$ in $\Bbb C$).

*There is an "absolute value" $|\cdot|$ that assigns to any $(x,y,z)\in\Bbb R^3$ a real number, its absolute value. Clearly this exists in $\Bbb R$, and also in $\Bbb C$ via $|x+yi|=(x^2+y^2)^{1/2}$. A natural property of the absolut value is that it interacts well with the multiplication of the system, that is $|uv|=|u| \cdot|v|$, and that it behaves as expected on the embedded $\Bbb R$, that is $|x|$ is the usual absolute value for $x\in\Bbb R$ when considering $x$ as an element of $V$.
One thing that already follows from this is that $|\cdot|$ is a norm on the real 3-dimensional vector space $V$, and one can define an inner product
$$\langle u,v\rangle := \frac12(|x+y|^2-|x|^2-|y|^2)$$
using the so-called polarization identities.
In particular, there must be an element $i\in V$ with  $\langle i,1\rangle=0$.
Then follows
$$\langle ix,x\rangle\> = \frac12(|ix+x|^2-|ix|^2-|x|^2) = \frac{|x|}2(|i+1|^2-|i|^2-|1|^2) = |x|\cdot\langle i,1\rangle = 0.$$
A: Short Answer: No,
A lot of responses bring up the fact the next closed algebraic set are quaternions. But these aren't perfect since there is no problem to which Quaternions naturally arise as a solution:
For example we know $i$ naturally is formed as the solution to the previously unsolvable $\sqrt{-1}$ that being said all polynomials have roots in the complex plane so we are guaranteed no other such  new algebraic unit (like) $i$ will be naturally formed.
Now I personally don't know of a theorem that says this CANNOT happen again. For example it could be that:
$$x^x = i$$
Or (if we define ${}^xx$ to mean tetration)
$${}^xx = i$$
Etc... following the pattern, may not have a solution in C. In that case we are now free to generate a new elementary unit $j$ however this elementary unit will be quite strange since it is associated with a higher operator so expressions such as:
$$j, j^2, j^3 ...$$
Are all unique and simplified, leading us to a now infinite dimensional system of numbers
So 3D is not possible, but I believe infinite dimensional is still possible.
Unless you like the unnatural creations that are the quaternions etc... (I only dislike them because of their lack of natural formation)
