Reduce over integers - why does MMA not give ~impossible?

Question

I know that Reduce is not Solve but from e.g. 212880 I also see that integer solutions may be found by it. However, I am clearly confused about what I should expect from it and why.

Consider this system of equations

$2^x = 3 k + 1 \land 2^{x-1} = 3 m + 1$

By inspection there are no solutions. If we put the question to MMA (11.0.1) like this

Block[{x, k, m}, Reduce[2^x == 3 k + 1 && 2^(x - 1) == 3 m + 1, {x, k, m}, Integers]] 

We obtain (via TeXForm)

$ (k|m|x)\in \mathbb{Z}\land k=\frac{2^x}{3}-\frac{1}{3}\land m=\frac{2^{x-1}}{3}-\frac{1}{3}$

Q1 Why does MMA not realise that the forms of $k, m$ are incompatible?

Q2 Why does MMA state "$ (k|m|x)\in Integers$" as Alternatives? The reduction is specified over the integers so they must all be integers simultaneously. Specifically, why is it not expressed as an And of each $\in$ Integers?

Q3 Summary: is it basically just saying that there is no reduction of the equations?

NB if I reduce over {x, k}, or {x, m} MMA returns False, which I might interpret as there being no solution, but why should I pick these variable pairs? Note that Reduce over {k, m} does not return False

UPDATE Summary of new & selected points from various responses.

From the documentation, the first line of Details and Options, says that the result of Reduce" always describes exactly the same mathematical set as expr.", and a further example further down confirms that False means there is no solution.

Ordering of variables is important: Reduce[..., {k, m, x}] = False i.e. it does give "~impossible".

Reduce might, like FindInstance be unable to provide an answer, but unlike the latter may not know that it can't and may not say so to the user.

From the documentation, $Element[x1|x2|...,dom]$ asserts that all the $x_i$ are elements of dom. (The justification for using this form to express is still wanting)

Note that $2^x$ is not integral for integer $x<0$

Answer 1

Reduce[2^x == 3 k + 1 && 2^(x - 1) == 3 m + 1, {x}, Integers]

False

If we using {x,k,m} as variable, MMA will try to give all the expression of all the variable {x,k,m},it maybe exceed the ability of MMA. If we only set {x}, MMA will refer {k,m} as parametric variables, this make it focus the relation between {x} and {k,m},that is we need to help MMA in this cases.

Answer 2

This can be done with Mathematica in such a way.

NMinimize[{(2^x - 3 k - 1)^2 + (2^(x - 1) - 3 m - 1)^2, 
x \[Element] Integers && m \[Element] Integers &&  k \[Element] Integers}, {x, m, k}]

{0.25, {x -> 0, m -> 0, k -> 0}}

The above result implies there is no solution of the system under consideration over the integers.

Addition. I think the result of (y=2^x)

FunctionConvexity[(y - 3 k - 1)^2 + (y/2 - 3 m - 1)^2, {y, k, m}]

1

guarantees the found minimum over the integers is global.

Addition 2.

NMinimize[{(y - 3 k - 1)^2 + (y/2 - 3 m - 1)^2, {y, k, m} \[Element]  Integers}, {y, k, m}]

{0.25,{y->1,k->0,m->0}}

implies {0.25, {x -> 0, m -> 0, k -> 0}} is the global minimum of (2^x - 3 k - 1)^2 + (2^(x - 1) - 3 m - 1)^2 over the integers since the global minimum of the convex function over the superset $\mathbb Z^3$ equals the minimum over the set $\{2^x|x\in \mathbb Z\}\times\mathbb Z^2$. Hope I am clear.

Addition 3. As @user293787 pointed out, $2^x$ is not an integer if $x$ is a negative integer. For this case

NMinimize[{(y - 3 k - 1)^2 + (y/2 - 3 m - 1)^2, {k, m} \[Element] 
Integers && y > 0 && y <= 1}, {y, k, m}]

{0.25, {y -> 1., k -> 0, m -> 0}}

completes the job.