The problem is to prove the following
If $\gcd(a,b) = c$, then $\gcd(a^m, b^m) = c^m$
I know that this can be solved easily by proving that $c\mid a \implies c^m \mid a^m$ and $c\mid b \implies c^m \mid b^m$. So the greatest common divisor of $a^m$ and $b^m$ is $c^m$. But the problem is that I want to use Bézout's Lemma (Identity) to prove this.
I first introduce the following notation:
$$a = ca' \quad \quad b = cb', \quad \quad \text{where $(a',b') = 1$.}$$
Now from Bézout's Lemma, becaus $a'$ and $b'$ are coprime integers then we have integer solutions for the following equation:
$$a'x + b'y = 1$$
It's obvious that also the following equation has an integer solution:
$$a'^mx_1 + b'^my_1 = 1$$
Although I'm not able to prove this using Bézout's Lemma.
Now multiply both sides by $c^m$ we get:
$$(a'c)^mx + (b'c)^my = c^m$$ $$a^mx + b^my = c^m$$
But Bézout's Lemma states that for given $c^m$ is the greatest common multiplier of $a^m$ and $b^m$ or it is a multiplier of it.
But how to prove that it's not a multiplier of the greatest common multiper, and in fact is the greatest common multiplier. Is it possible to solve this problem with Bézout's Lemma?
You need both parts of your work to get the proof.
Your argument "this can be solved easily" only amounts to a proof that that $c^m$ divides the GCD: it doesn't actually prove $c^m$ actually is the GCD.
The rest of your work provides the other half: that the GCD divides $c^m$.
The only way I see to do the first half of the proof with Bezout's lemma doesn't really amount to anything different: it observes that $c^m | a^m$ and $c^m | b^m$, and thus $c^m | a^m x + b^m y$ for all $x,y$.
Here's a useful trick to fill in your gap: if $c x + d y = 1$, then we also have $(c x + d y)^{2m-1} = 1$ (any power at least $2m-1$ would suffice for this trick). You can expand this and collect terms to find an identity $c^m u + d^m v = 1$.