Why are there no positive rational solutions for the following Diophantine equation over the field of rational numbers?

Question

Please consider the following Diophantine equation for any integer, m > 5:

$$(k + λ)^{m-1}  +  (k + λ)^{m-2} \cdot k  +  . . .  +  (k + λ) \cdot k^{m-2}  +  k^{m-1} =   4^{m} /  λ$$

where the rational number,  λ,  is such that $$0 <  λ <  4^{m}.$$ 

Applying Galois theory, can anyone explain why there are no positive rational solutions in k for the following irreducible (over the set of rational numbers) polynomial equations of degree, m ≥ 6, whose real solutions, k, are probably irrational numbers or transcendental numbers?

Please do not apply Fermat's Last Theorem (FLT).

If m = 6, we have the following equation: $$6λ\cdot k^{5}+15λ^{2}\cdot k^{4} + 20λ^{3}\cdot k^{3} + 15λ^{4}\cdot k^{2} + 6λ^{5}\cdot k + λ^{6} = 4096$$ with $$0 <  λ <  4096.$$

If m = 7, we have the following equation: $$7λ\cdot k^{6}+21λ^{2}\cdot k^{5} + 35λ^{3}\cdot k^{4} + 35λ^{4}\cdot k^{3} + 21λ^{5}\cdot k^{2} + 7λ^{6}\cdot k + λ^{7} = 16384$$ with $$0 <  λ <  16384.$$

Reference: Wolfram Cloud (https://www.wolframcloud.com/) code to generate the above equations for values, m=6 and m=7:

f[m_]:= (n=m;l = k+[Lambda] ; Return[{Together[[Lambda]*(l^n-k^n)/(l-k)],4^n}]).

Reference link: Prove a statement about a conditional Diophantine equation.

Answer 1

[ EDIT ]  The answer below was posted before the question was edited to specifically ask for Galois proofs not using FLT. From the references that were added, the question appears to be related to an attempt at an independent proof of FLT.  While the answer no longer applies to the question as restated, I am leaving it here just so that others are not tempted to repost the same obvious answer.

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Multiplying both sides by $(k+\lambda)-k=\lambda$ gives:

$$(k+\lambda)^m - k^m = 4^m \quad \iff \quad k^m + 4^m = (k+\lambda)^m$$

By FLT the latter has no non-trivial rational solutions for $\;m \gt 2\;$ and the condition $k \gt 0$ excludes the trivial solutions $\;k=0, \lambda=4\;$ and (for odd $m$) $\;k=-4,\lambda=4\;$.

As a side note, the premise $0 \lt \lambda \lt 4^m$ is not necessary for the conclusion to hold, and $m \gt 2$ is sufficient (rather than $m \gt 5$).