Jelly, 5 bytes

_2&’$

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Uses the fact that a given $n$ returns $0$ iff $n-2$ is a power of $2$, as pointed out by Kevin Cruijssen and the n-2 & n-3 trick

How they work

_2&’$ - Main link. Takes n on the left
_2    - n-2
    $ - To n-2:
   ’  -   Decrement; n-3
  &   -   n-2 & n-3

Husk, 6 bytes

¬εΣḋ≠2

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This one got simplified really quickly.

APL (Dyalog Extended), 9 bytes

⊃∧/⊤⎕-2 3

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A full program that takes a single number $n$ from stdin, and prints 1 for true, 0 otherwise. APL doesn't have bitwise functions, so we need to explicitly convert to binary and apply boolean functions on each bit.

How it works

⊃∧/⊤⎕-2 3  ⍝ Input: n (from stdin)
    ⎕-2 3  ⍝ [n-2, n-3]
   ⊤       ⍝ Convert to binary
           ⍝ (each number becomes a column in a matrix, aligned to bottom)
⊃∧/        ⍝ Check if the MSB of both numbers are 1,
           ⍝   i.e. the bit lengths of the two are the same

Charcoal, 8 bytes

‹¹Σ⍘⊖⊖Ｎ²

Try it online! Link is to verbose version of code. Outputs a Charcoal boolean, i.e. - if RLD numbers exist otherwise no output. Explanation:

      Ｎ     Input as a number
    ⊖⊖      Decremented twice
   ⍘   ²    Converted to base 2
  Σ         Digital sum
‹¹          Is greater than 1

The only binary numbers with a digital sum of 1 or less are 0 and powers of 2, so by @KevinCruijssen's proof a solution exists for all other values of n.

Retina 0.8.2, 21 bytes

.+
$*
^11(1(11)+)1*$

Try it online! Link includes test cases. Explanation: The first stage converts to unary, while the last stage uses @KevinCruijssen's observation that a solution exists if n-2 has a nontrivial odd factor.

GolfScript, 5 bytes

In GolfScript 0 is falsy while any other value is truthy.

2-.(&

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JavaScript (Node.js), 10 bytes

In JS 0 is falsy and everything else is truthy. Again, another port of Kevin Cruijssen's answer!

x=>x-2&x-3

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Ruby, 14 bytes

Port of Kevin Cruijssen's answer, remember to upvote them!

->x{x-2&x-3>0}

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Bash, ²² 20 bytes

Uses Kevin Cruijssen's formula.

Returns $1$ for falsy and $0$ for truthy.

Saved 2 bytes thanks to dingledooper!!!

echo $[!($1-2&$1-3)]

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C (gcc), 19 bytes

Uses Kevin Cruijssen's formula.

Returns $1$ for falsy and $0$ for truthy.

f(n){n=!(n-2&n-3);}

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Python 3, ¹⁹ 18 bytes

Uses Kevin Cruijssen's formula.

Returns True/False.

Saved a byte thanks to dingledooper!!!

lambda n:n-2&n-3>0

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05AB1E, 6 5 bytes

ÍD<&Ā

-1 byte thanks to @xnor and @Noodle9.

Try it online or verify the first $[2,100]$ test cases.

Explanation:

Í       # Decrease the (implicit) input-integer by 2
        # Check that this input-2 is a power of 2 by:
 D      #  Duplicating it
  <     #  Decrease the copy by 1 (so integer-3)
   &    #  Take the bitwise-AND of input-2 and input-3
    Ā   #  Check that this is NOT 0
        # (after which the result is output implicitly)

But wait, I don't see any use of bases nor rotation!

When I saw the challenge in the Sandbox and was working on a solution, I noticed that the only falsey values in the first $n=[2,500]$ bases formed the sequence A056469: number of elements in the continued fraction for $sum_{k=0}^n (frac{1}{2})^{2^k}$, which could be simplified to $a(n)=leftlfloor2^{n-1}+2rightrfloor$. Here a copy of the first 25 numbers in that sequence as reference:

2, 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026, 2050, 4098, 8194, 16386, 32770, 65538, 131074, 262146, 524290, 1048578, 2097154, 4194306, 8388610

It can also be note that all the numbers in this sequence are of the form $a(n)=2^n+2$, so checking whether $n-2$ is a power of $2$ will verify whether it's in this sequence. Since we want to do the invert here, and having a falsey result if it's in this sequence (or truthy if it's NOT in this sequence), we'll do just that, resulting in the code above.

Mathematical proof that all falsey cases of the Left-Rotate-Double numbers are of the form $2^n+2$:

Quote from @saulspatz at the Math SE, who provided me with this Mathematical proof to back up my theory I based on the first $n=[2,500]$ test cases. So all credit for this proof goes to him/her.

If $m$ is a $(d+1)$-digit Rotate-Left-Double number in base $n$, then $$m=xn^d+ytag1$$ where $dgeq1, 0<x<n, 0leq y<n^d$. (Includes the rule that the number can't start with $0$.) Rotating $m$ gives $ny+x$, so we have $2xn^d+2y=ny+x$ or $$(n-2)y=(2n^d-1)xtag2$$ If $n=2^k+2$ then $(2)$ gives $(n-2)|x$ (which means $x$ is divisible by $(n-2)$), since $2n^s-1$ is odd. But then $ygeq 2n^d-1$ which contradicts $y<n^d$.

To show that these are the only falsey numbers, let $p$ be an odd prime dividing $n-2$. (Such a $p$ exists because $n-2$ is not a power of $2$.) In $(2)$ we can take $x=frac{n-2}p<n$ and we have to show that there exist an exponent $d>0$ and $0leq y<n^d$ such that $$py = 2n^d-1$$ If we can find a $d$ such that $p|(2n^d-1)$, we are done, for we can take $y = frac{2n^d-1}p<n^d$.

By assumption, $n-2equiv0pmod{p}$ so $nequiv 2pmod p$. Therefore, $$2n^dequiv1iff 2cdot2^dequiv1 iff 2^{d+1}equiv 1pmod p,$$ and by Fermat's little theorem, which states that $a^{p-1}equiv 1pmod p$, we can take $d=p-2$, because $$2^{p-2+1}equiv 1 iff 2^{p-1}equiv 1 pmod p$$

This completes the proof.