Prove not context free
For large enough $p$, consider the word $w = 0^{p+1}1^{p+1}\#0^{p+1}1^{p+2} \in L$. Mark the part $1^{p+1}\#0^{p+1}$. According to Ogden's lemma, we can write $w = uxyzv$ such that $xyz$ contains at least $1$ and at most $p$ marked symbols, and $ux^iyz^iv \in L$ for all $i \geq 0$. The pumped part $xyz$ cannot lie all to the right of $\#$ since then pumping it down would result in a word not in the language (here we crucially use the fact that only the $0^{p+1}$ part to the right of $\#$ is marked). It also cannot lie all to the left of $\#$ since then pumping it up would result in a word not in the language. It follows that the part of $xyz$ to the right of $\#$ is of the form $0^k$, and the part of $xyz$ to the left of $\#$ is of the form $1^\ell$ (otherwise, there would be more than $p$ marked symbols). However, pumping up, the resulting word is not in the language. This contradiction shows that $L$ is not context-free.