Language of lists of words, not all of which are different, is not context-free
If $L$ were context-free then so would $L' = d(L \cap (0+1)^*\#(0+1)^*)$ be, where $d$ is the homomorphism that deletes $\#$. However, $L'$ is the language of squares (words of the form $w^2$), which is well-known not to be context-free.
If for some reason you have to prove that $L$ is not context-free directly using the pumping lemma, this suggests looking at the proof that $L'$ is not context-free and trying to adapt it.