Euler's Formula is as follows;
$$e^{i\theta} = \cos\theta + i \sin \theta.$$
You can think of the number $1$ as $e^{\ln1}$ and hence of $1^i$ as $e^{i\ln1}$. But $\ln 1 = 0,$ so really you have $e^{0i} = e^0 = 1$, as I'm sure you're aware. Even if we look at this by feeding $0$ into Euler's Formula, we have
$$e^{0i} = \cos(0) + i\sin(0) = 1.$$
In principle we could do what you suggest -- take $\mathbb R^2$ and associate every point $(x,y)$ to the number $x+13k$. Though the trouble with that particular plan is that each number now represents many different points -- for example, $(13,0)$ and $(0,1)$ and $(26,-1)$ are now all associated to the number $13$. This means that we can't use the scheme for anything where we calculate a number and that number points to exactly one point in the plane.
We could, however, do something more general. Take some field that extends $\mathbb R$, pick some element $\alpha$ in that field, and then represent $(x,y)\in\mathbb R^2$ by $x+\alpha y$.
As it went for $13$, if we pick $\alpha\in\mathbb R$, then we get something where a number doesn't represent a unique point. Suppose, however, that we steer clear of that case, and furthermore that we end up in the lucky situation that every element of the field represents some $(x,y)$ in the plane.
Something wonderful happens then -- namely, we can then prove (though not in the space left for me in this margin) that the field we're using must be isomorphic to $\mathbb C$ -- in other words the field is essentially the complex numbers, just called something different. In particular, somewhere in the plane there is an $(x,y)$ whose corresponding number behaves exactly like $i$.
So we could actually have said: Pick some complex number $\alpha$, and let $(x,y)$ correspond to $x+\alpha y$. As long as $\alpha$ is not real, this will give us a perfectly good one-to-one correspondence between points and complex numbers.
Now, among all the possible choices of $\alpha$ it turns on that exactly when $\alpha=i$ or $\alpha=-i$ we get the additional nice property that multiplication by any fixed nonzero complex number will correspond to a transformation of the plane that takes geometric figures to similar geometric figures.
Having multiplication correspond to similarity transforms is a pretty nifty property, which is a reason to prefer the representation $x+iy$ over other possible $x+\alpha y$.
Best Answer
One way to think of the complex plane $\mathbb{C}$ is to think of it as the real coordinate (or $xy$-) plane $\mathbb{R}^2$. In $\mathbb{R}^2$, we have two real-numbered axes (in particular, the $x$ and $y$ axes). An element of $\mathbb{R}^2$ is an ordered pair $(x, y)$, where $x$ and $y$ range over all the real numbers $\mathbb{R}$.
What I am about to say is not exactly rigorous, but I purposefully omit rigor in favor of your understanding.
Now, what happens if we decide to multiply $y$ by $i$, for all $y$ in $\mathbb{R}$? We obtain $i \mathbb{R}$, where each element is of the form of $i y$, where $i$ is of course defined as usual and $y$ is a real number. Consequently our vertical $y$-axis now becomes imaginary, and we obtain the complex plane $\mathbb{C}$. Note that just like the real plane $\mathbb{R}^2$, $\mathbb{C}$ is 2-dimensional. To recap, $\mathbb{C}$ has two axes, one real axis and one imaginary axis.
What does an element of $\mathbb{C}$ look like? An element $z$ of $\mathbb{C}$ has the form $a + bi$, where $a$ and $b$ are real numbers; we call $z$ a complex number. If we were to graph a complex number $z = a + bi$, we would graph it just like we would an ordered pair $(a, b)$ in $\mathbb{R}^2$. E.g., the number $z = 1 + i$ would correspond to the point $(1, 1)$; the number $z = 1$ would correspond to the point $(1, 0)$. This second example demonstrates that a real number is in fact a complex number as well.
I am not sure if you have seen linear algebra or not, but your statement in the second sentence of your first paragraph can be explained with such ideas. In the real plane $\mathbb{R}^2$, we have two directions; namely, the $x$ and $y$ directions. It turns out that we only need the number $x = 1$ to generate the entire $x$-axis. Similarly we may do the same with the $y$-axis. Together, we obtain the whole plane $\mathbb{R}^2$. $\mathbb{C}$ carries the same idea, except your vertical axis is generated by $i = 1\cdot i$ instead of just $1$.
To further address your question, recall $i$ is defined as $i = \sqrt{-1}$. Thus, $i^2 = -1$, $i^3 = -i$ and $i^4 = 1$. What have we accomplished here? We have obtained directions for the vertical and horizontal axes. From here we may generate all of $\mathbb{C}$.
I am not entirely sure how to answer your next few questions, but recall from above that $\mathbb{C}$ is 2-dimensional. Thus, $\mathbb{C}^2 = \mathbb{C} \times \mathbb{C}$ would be 4-dimensional. Generalizing this notion, $\mathbb{C}^n$ is $2n$-dimensional. Hence I do not think there is really an analogy to the $z$ axis if we are referring to the third axis in $\mathbb{R}^3$. The best analogy I can come up with are several-dimensional complex number systems such as $\mathbb{C}^2$.