Proving that $L = \{a^m b^n | m \% n = 0 \}$ is not context-free

Question

For language $L = \{a^m\, b^n\: |\: m \:\%\: n = 0 \}$, that is, $m$ is a multiple of $n$, I'm trying to find a proof that it isn't a context free. I know it isn't regular, but it also doesn't seem to be context-free to me. However, I am not sure how I would go about proving that it isn't context-free. My quick attempt follows:

First, assume to contrary that $L$ is context-free. Let $p$ be the guaranteed pumping length. Choose string $s\,=\,a^{\,p+1}\,b^{\,p+1}$, where $s\,\in\,L$. By the pumping lemma, we can write $s$ as $s\,=\,uxwvy$.

If $vx$ contains either $a's$ or $b's$, then $uv^2wx^2y\:\not\in\:L$.

If $vx$ contains both $a's$ and $b's$, then $uv^2wx^2y\:\not\in\:L$.

Therefore, we have a contradiction and $L$ cannot be context-free.

However, I don't believe I have the proof completely right, and I would like an elaboration how this proof could be better presented.

EDIT: Is this language context-free? Why or why not?

We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. — Raphael, Oct 18 '16 at 09:29
Note our reference question which has a good explanation with examples. — Raphael, Oct 18 '16 at 09:30
Your second "contradiction" isn't necessarily a contradiction. What if $x=a^k$ and $v=b^k$? — Rick Decker, Oct 21 '16 at 16:33

Proving that $L = \{a^m b^n | m \% n = 0 \}$ is not context-free

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