How many $6$-sided and $8$-sided standard dice exist?

For six-sided there are just two possibilities. This is because the $1$, $2$ and $3$ have to meet at a corner and once you place them, the rest of the die is determined. You can either have $1,2,3$ going clockwise or anticlockwise round the corner, leading to two possibilities which are mirror images of each other.

For eight-sided it is much more complicated, as there are many ways to arrange $1,2,3,4$ without having any of them opposite another. If $1$ and $2$ are adjacent (i.e. on two faces which share an edge), there are $4$ choices for where to place $3$ and for each of those there are $2$ choices for which face is $4$ (then $5,6,7,8$ are determined). This gives $8$ different options, and there are $8$ more obtained by swapping $2$ and $7$ so that $2$ and $1$ are not adjacent. So in total there are $16$ arrangements.

(edit) Incidentally, the total number of ways to number a die with $2n$ faces is the number of standard ways times $(2n-1)!!$ (i.e. the product of all odd numbers up to $(2n-1)$). This is because there are $(2n-1)!!$ ways to divide the numbers into $n$ pairs, and then the number of ways to put each pair on opposite faces is the same as the original problem. So there are $30$ six-sided dice and $1680$ eight-sided dice.

Here is another way of counting the octahedral (8=sided) case.

Bring the triangle with $1$ on to the front. It has three numbers adjacent from the pairs $(2,7);(3,6);(4,5)$. Rotate so that $2$ or $7$ is at the base of the triangle ($2$ possibilities). For each case there are $4$ possibilities for the left side of $1$ and two possibilities remain for the right-hand side. Total $16$.

For the cube. I can always rotate so I have $1$ at the front and $2$ at the base. Then there are two possibilities for placing $3$ and $4$.

Another way to do the counting:

Let's start by counting how many ways we can put numbers $1$ through $2n$ on a regular $2n$-sided die if we fix the position of the die in space (do not allow rotating the die in order to change where the numbers appear). This is just the number of ways we can permute the $2n$ numbers among the $2n$ places where they can be written; the number of ways is $$(2n)! = (2n)(2n-1)(2n-2)\cdots(3)(2)(1).$$

Now count the number of ways we can orient the die (allowing it to rotate now) so that it occupies the same space as before, but not necessarily with the same numbers facing the same directions. Call this number $|R_{2n}|.$ This is the number of times each possible numbering of the die was repeated when we numbered the faces without considering the possibility of turning the die. So the total number of ways to number a regular $2n$-sided die, not counting rotations of the same die as different numberings, is $$\frac{(2n)!}{|R_{2n}|}.$$

But as pointed out in another answer, the total number of ways to number a regular $2n$-sided die is exactly $(2n-1)!! = (2n-1)(2n-3)(2n-5)\cdots(5)(3)(1)$ times the number of standard numberings of the die (if any standard numberings exist). So the total number of standard numberings is $$\frac{(2n)!}{(2n-1)!!\cdot|R_{2n}|} = \frac{(2n)!!}{|R_{2n}|},$$ where $(2n)!! = (2n)(2n-2)(2n-4)\cdots(6)(4)(2).$

There are only five regular polyhedra, for which these numbers come out to: \begin{array}{rrrrrr} 2n & (2n)! & |R_{2n}| & \text{number of dice} & (2n)!! & \text{standard}\\ \hline 4 & 24 & 12 & 2 & 8 & - \\ 6 & 720 & 24 & 30 & 48 & 2 \\ 8 & 40320 & 24 & 1680 & 384 & 16 \\ 12 & 479001600 & 60 & 7983360 & 46080 & 768 \\ 20 & 2432902008176640000 & 60 & 40548366802944000 & 3715891200 & 61931520 \end{array}

There obviously are no "standard" numberings for a $4$-sided die, since no face has an "opposite" face, but the formulas still count how many ways the die can be numbered without that restriction.