Question: I know that the the set of real numbers has been proved uncountable by mathematicians, so my question is why this is wrong.  

In countability arguments that I have seen the numbers are laid out in a grid with 1,2,3,4,... as rows and columns and the number at the nth row and mth column is n/m. So what if we have an extra dimension k, that represents the power, so at depth 3 the power is 1/3  

so it becomes $ ({\frac{n}{m}})^{(\frac{1}{k})} $.  

It should be possible to visit every position in the cube by a similar procedure akin to the one used for 2 dimensional grid.  

Any non fractional power will give another rational number so only 1/k is actually necessary. While this only appears to work for the solution to a polynomial, such solutions are not rational numbers. Which would make some sets of irrational numbers countable, but that doesn't quite make sense to me.

So therefore there must be something wrong with my ideas but I don't know what it is.

Answer: 
As you point out, every number appearing in your cube is a root of a polynomial with rational coefficients, so it is an algebraic number. And the set of algebraic numbers over $\mathbb{Q}$ is countable (for example, see [Prove that the set of all algebraic numbers is countable](https://math.stackexchange.com/questions/489186/prove-that-the-set-of-all-algebraic-numbers-is-countable)).
You have only found a countable set which includes SOME irrational numbers. How do you know ALL of them can be written in that way?

What I am saying is, there are some irrational numbers ($\pi$ or any other transcendental number, for example), which cannot be written as $q^{1/k}$ for some rational number $q$ and integer $k$.