Ternary and decimal expansion of an integer solely consists of $2$ and $0$.

Each digits of the decimal expansion of the integer $2022$ (this year) consists of $0$ or $2$ and also, each digits of the ternary expansion of the same integer $2022$ (which is $2202220_3$) consists of $0$ or $2$. I wonder if there are infinitely many such integers.

I think they will run out after a while. This is not a proof, just heuristics.
There are $2^n$ of these numbers of length $n+1$. One approach is to think of the base 3 version as a random set of digits. There will be about $(n+1)\log_{3}10$ base 3 digits, each has $2/3$ chance of being 0 or 2.
The chance of any of the $n+1$ digit decimal numbers being just 0s and 2s in ternary would be $$\approx C\left(2\left(2\over3\right)^{\log_310}\right)^n\\ \approx C(0.855^n)$$ This has a finite sum, so I expect finitely many of these numbers.

I am writing this as an answer rather than as a comment because the formatting is better. In a computer search I found the following thirteen values that satisfy the conditions (all values are given in decimal):

$0\\ 2\\ 20\\ 222\\ 2000\\ 2022\\ 22220202002202\\ 22220202002220\\ 22220202002222\\ 22220202020022\\ 22220202020202\\ 2200222000020222022200\\ 2200222000020222022202$

I extended the search up to (approximately) $2.02\times 10^{27}$ and found nothing further.