Is $2048$ the highest power of $2$ with all even digits (base ten)?

This seems to be similar to (I'd venture to say as hard as) a problem of Erdős open since 1979, that the base-3 representation of $2^n$ contains a 2 for all $n>8$.

Here is a paper by Lagarias that addresses the ternary problem, and for the most part I think would generalize to the question at hand (we're also looking for the intersection of iterates of $x\rightarrow 2x$ with a Cantor set). Unfortunately it does not resolve the problem.

But Conjecture 2' (from Furstenberg 1970) in the linked paper suggests a stronger result, that every $2^n$ for $n$ large enough will have a 1 in the decimal representation. Though it doesn't quantify "large enough" (so even if proved wouldn't promise that 2048 is the largest all-even decimal), it looks like it might be true for all $n>91$ (I checked up to $n=10^6$).

This sequence is known to the OEIS.

Here are the notes, which give no explicit answer but suppose that your conjecture is correct:

Are there any more terms in this sequence?

Evidence that the sequence may be finite, from Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jun 23 2002:

1) The sequence of last two digits of $2^n$, A000855 of period $20$, makes clear that $2^n > 4$ must have $n = 3, 6, 10, 11,$ or $19 (\text{mod }20)$ for $2^n$ to be a member of this sequence. Otherwise, either the tens digit (in $10$ cases), as seen directly, or the hundreds digit, in the $5$ cases receiving a carry from the previous power's tens digit $\geq 5$, must be odd.

2) No additional term has been found for n up to $50000$.

3) Furthermore, again for each n up to $50000$, examining $2^n$'s digits leftward from the rightmost but only until an odd digit was found, it was only once necessary to search even to the 18th digit. This occurred for $2^{12106}$ whose last digits are $\ldots 3833483966860466862424064$. Note that $2^{12106}$ has $3645$ digits. (The clear runner-up, $2^{34966}$, a $10526$-digit number, required searching only to the $15$th digit. Exponents for which only the $14$th digit was reached were only $590, 3490, 8426, 16223, 27771, 48966$ and $49519$ - representing each congruence above.)

Here's a heuristic justification that none exist.

The odds of a uniformly-randomly chosen $n$-digit number has all of its digits even is $$ p(n) = \frac{5}{9} \frac{1}{2}^{n-1} $$

(note that your last digit data supports this method of analysis: roughly $1/2^n$ of the last $n$-digits are all-even)

On average, there are $\log_2(10)$ $n$-digit numbers in the sequence of powers of 2.

So, at random, the odds that there does not exist a power of 2 of $k$ or more digits that are all even is $$ \prod_{n=k}^{\infty} \left(1 - \frac{5}{9} \frac{1}{2^{n-1}}\right)^{\log_2 10}$$ For $k=10$, wolfram alpha computes this to be $0.992814$. So there's a less than 1% chance of there being another power of 2 whose digits are all even.

Of course, after you exhaust the small ones to see one doesn't exist, you can plug in much larger values into $k$....