Is this number an integer power of -2? There are clever ways of determining whether a number is a power of 2.  That's no longer an interesting problem, so let's determine whether a given integer is an integer power of -2.  For example:

-2 =&gt; yes: (-2)¹
-1 =&gt; no
0 =&gt; no
1 =&gt; yes: (-2)⁰
2 =&gt; no
3 =&gt; no
4 =&gt; yes: (-2)²


Rules


You may write a program or a function and use any of the standard methods of receiving input and providing output.
Your input is a single integer, and output must be a truthy value if the integer is an integer power of -2, and a falsy value otherwise.  No other output (e.g. warning messages) is permitted.
The usual integer overflow rules apply: your solution must be able to work for arbitrarily large integers in a hypothetical (or perhaps real) version of your language in which all integers are unbounded by default, but if your program fails in practice due to the implementation not supporting integers that large, that doesn't invalidate the solution.
You may use any programming language, but note that these loopholes are forbidden by default.


Winning condition

This is a code-golf contest: the answer which has the fewest bytes (in your chosen encoding) is the winner.
Mathematica, 22 bytes

EvenQ@Log2@Max[#,-2#]&amp;


Try it online! (Using Mathics instead, where this solution also works.)

I tried to find a solution with bitwise operators for a while, and while one definitely exists, I ended up finding something which is probably simpler:


Max[#,-2#] multiplies the input by -2 if it's negative. Multiplying by another factor of -2 doesn't change whether the value is a power of -2 or not. But now all odd powers of -2 have been turned into even powers of -2.
But even powers of -2 are also even powers of 2, so we can use a simple Log2@... and check if the result is an integer (to check whether it's a power of 2). This already saves two bytes over Log[4,...] (another way to look at even powers of -2).
As an added bonus, checking whether a value is an even integer is shorter than just checking whether it's an integer: we can save three more bytes by using EvenQ instead of IntegerQ.