A context free grammar proof
Hint...

There may be a direct application of the pumping lemma, but I suggest you take a look at closure properties. Consider the non-CFL in the first answer to the reference post, $P=\{a^p:p \text{ prime}\}$, which is clearly a cousin of $L$. Can you find a (fairly simple) operation on $L$ that preserves CFLs and more or less yields $P$?  If so, you are done, except perhaps to tidy up with some other operations to deal with special cases for $P$.

In general, when languages are relatively "dense", that is, have a high proportion of members to non-members for all given lengths, it's harder to apply pumping arguments, because it's more work to pump to get outside the "dense" set; in fact, sometimes it's impossible.  $P$ is a nice "sparse" set, so pumping works well, as shown in the reference post.  $L$ is quite a bit "denser", so transforming it to a "sparser" form is a good tactic to try.

The classical example of this principle (for non-regular languages) is the very dense set $\{a^ib^j:i \neq j\}$ and the corresponding sparse set is $\{a^ib^i\}$. The operation in this case is complement, again with a "cleanup" operation needed as well.  As part of the above hint, the operation there is not complement, which doesn't work for CFLs anyway.