How exactly do quantum numbers increase in relation to energy levels in more than one dimension? When you have increasing energy levels in 2 or 3 dimensions how does the values of quantum numbers $n$ increase for each dimension?
For example if you have ground state $E_1$ then you have for $(n_x,n_y,n_z)$ is $(1,1,1)$ in 3 dimensions.
So does this mean for energy $E_2$ you have $(2,2,2)$ ?
I've only learnt it in 1 dimension so i don't understand how it works in higher dimensions?
 A: For a particle in a box (also known as a particle in an infinite potential well) in $d$ dimensions, the Hamiltonian inside the box is given by $$\hat{H}=\frac{1}{2m}\sum_{i=1}^d{p}^2_i$$where $p_i$ is the momentum operator in the $i^{\text{th}}$ direction. As you already know (I infer so from your question) that using the boundary conditions, we arrive at the conclusion that $p_i$ must be of the form $$p_i=\frac{n_i\pi}{L_i}$$where $L_i$ is the length of the box in the $i^{\text{th}}$ direction and $n_i\in\mathbb{N}$. Thus, the energy of an eigenstate is given by $$E=\frac{\pi^2}{2m}\sum_{i=1}^d\frac{n_i^2}{L_i^2}$$.
Now, in the case of one--dimensional box, $E$ only depends on $n_x$ and is monotonically increasing with $n_x$. This means that the spectrum of a one--dimensional particle in a box is non--degenerate, and the value of the energy depends only on one quantum number, namely, $n_{x}$. One can simply write that $E=E_{n_x}$
However, in the case of a multidimensional box, $E$ depends on the multiple quantum numbers, namely, $\{n_i\}$ where $i>1$. And, we have to write $E=E_{\{n_i\}}$ to signify that $E$ depends on multiple quantum numbers.
Moreoever, multiple different sets of values of $n_i$ can produce the same value of energy. For example, let's consider the case where $L_i=L$ and $i=3$. All the following configurations of $n_i$ produce the same value of energy:$$n_x=2,n_y=1,n_z=1$$ $$n_x=1,n_y=2,n_z=1$$ $$n_x=1,n_y=1,n_z=2$$
Thus, we see that the spectrum of a particle in a box can be degenerate in a higher dimension.
So, to answer your question, the first excited energy state in a three-dimensional particle in a box (with equal sides) can correspond to all of the above configurations of quantum numbers.
A: 
So does this mean for energy $E_2$ you have $(2,2,2)$?

This is wrong! The 3-Dimensional box (and 2-Dimensional) have degeneracy is energy state that there 2 or more level for which the energy is same. Like for $E_2$, We have $(2,1,1),(1,2,1),(1,1,2)$.
Note that for 3-D box, The energy given by
$$E_{n_1,n_2,n_3}=(n^2_1+n^2_2+n^2_3)\frac{\pi^2\hbar^2}{2mL^2}$$
