Are there numbers that describe themselves in some base but not according to the patttern $6210001000$?
For bases sufficiently large (as @Goos noted, there are examples for low bases), it is impossible. I will use the facts that there must be as many numbers as the base, and that the digits must sum to the base (since they count the digits in the number).
Take the base $=n$, and consider the first digit $k$. $k$ cannot be zero, because that would say there are zero zero's, which there wouldn't be! So there are $k$ zero's, and at least one $k$ (for now), meaning we have $$n-k-1$$ digits left to fill. These digits must add to the remaining $n-k-k\cdot 0=n-k$, and the only $n-k-1$ digits which sum to $n-k$ are $$1,1,\ldots,1,2$$ (we have $n-k-2$ one's). I will eliminate $k=1,2$ later. So we have $2$ of something, not $k$'s and not zero's. We must have $2$ one's. And then we have one $k$ and one $2$, which works. The condition that the numbers sum to the base gives $$k+1+1+2+k\cdot 0=k+4=n \implies k=n-4$$ so the only possible self-descriptive number is $$(n-4)(2)(1)0\ldots(1)0\ldots0.$$
If $k=1$, then there will have to be two one's, and one two, and one zero. This gives the only possibility $$(1210)_4$$ in base $4$.
If $k=2$, then there must be two two's. We have two zero's, so there are one or no one's. The two possibilities here are $$(21200)_5,\quad (2020)_4$$ So we have enumerated all self-descriptive numbers.
Yes, there are, but it's not known if there are a lot of them.
Let me rephrase "the pattern 6210001000" algebraically like so: $(b - 4)b^{b - 1} + 2b^{b - 2} + b^{b - 3} + b^3$. So the smallest base in which this results in a digit pattern similar to 6210001000 is base 7. But since $6^5 = 7776$, maybe we can do an exhaustive search in binary, ternary, quartal, quintal and heximal.
This turns up 1210 in quartal, which is 100 in decimal, 2020 in quartal, which is 136 in decimal, and 21200 which is 1425 in quintal. None of these are 292 nor 901, so they are not of the described pattern. There are apparently none in base 6.
Not a complete answer to your question, but I believe that the problem you're describing can be reduced to the following:
Given a natural number $B>5$, prove that there is only a single multi-set $M$ of natural numbers, such that $\sum{M}=\prod{M}$ and $|M|=B-4$.
For example:
- $B=10$
- $M=\{6,2,1,1,1,1\}$
- $\sum{M}=\prod{M}$ and $|M|=B-4$
If there is more than a single such multi-set, then there are numbers that describe themselves in some base but not according to the illustrated pattern.