Finding bicliques in a bipartite graph of minimum size
Here is an algorithm with $O(n^{2+\lg k})$ running time.  It's not polynomial in $k$ and $n$, but for fixed $n$ it is polynomial in $k$, and for fixed $k$ it is polynomial in $n$.  Here is the algorithm:


Let $\gamma = \lg k$.
Find all bicliques such that $|U'| \le \gamma$ and $|V'| \ge \delta$.  Output $k$ of them (stop early once you have found $k$ of them).


In step 2, it's easy to find all bicliques such that $|U'| \le \gamma$, $|V'| \ge \delta$ in $O(n^{2+\gamma})$ time: just enumerate all possibilities for $U'$.  By the definition of $\gamma$, this takes $O(n^{2+\lg k})$ time, since there are only ${|U| \choose 0 } + \dots + {|U| \choose \gamma} = O(|U|^{\gamma}) = O(n^{\gamma}) = O(n^{\lg k})$ possible subsets $U'$ of size $\le \gamma$.

But will step 2 find enough bicliques?  You might worry that it will find fewer than $k$ bicliques, even when there are at least $k$ in the graph.  Not to worry; that can't happen.  In particular, I'll show that if there is a biclique that is too large to be found in step 2, then there are at least $k$ bicliques that are small enough to be found in step 2, so step 2 will find enough to output $k$.  The only way that step 2 can fail to output $k$ bicliques is if the graph does not contain $k$ bicliques.

Let's prove that claim.  If step 2 outputs fewer than $k$ bicliques and there exists a biclique that the algorithm didn't output, that means there exists a biclique with $|U'| \ge \gamma+1$ and $|V'| \ge \delta$.  What are the consequences of this?  Well, there are two cases, depending on the size of $U'$:


Case 1. Suppose $\gamma+1 \le |U'| \le 2\gamma$.  Then you can find at least $2^{\gamma}$ more such that $|U'| \le \gamma$ and $|V'| \ge \delta$.  Why?  Well, take any subset of $U'$ of size at most $\gamma$, and keep $V'$ unchanged.  Then this will give a new biclique, and it will meet all the conditions of the original problem.  The number of such subsets is ${|U'| \choose 0} + {|U'| \choose 1} + \dots + {|U'| \choose \gamma}$, which is at least $2^{U'}/2$ (since $\gamma \ge |U'|/2$).  Now $2^{U'}/2 \ge 2^{\gamma+1}/2 \ge 2^{\gamma} \ge k$, so this gives us $k$ more bicliques such that $|U'| \le \gamma$ and $|V'| \ge \delta$.  Thus in this case there exist enough bicliques that step 2 can find $k$ of them.
Case 2. Suppose $|U'| \ge 2\gamma$.  Then you can find at least $2^{\gamma}$ more bicliques such that $|U'| \le \gamma$ and $|V'| \ge \delta$.  Why?  Again, take any subset of $U'$ of size at most $\gamma$; that will lead to another biclique that can be found in step 2.  How many such subsets are there?  There are at least ${|U'| \choose \gamma}$ of them, and  ${|U'| \choose \gamma} \ge (|U'|/\gamma)^{\gamma}$ more (see the first bound here).  By assumption, $|U'|/\gamma \ge 2$, so this means there are at least $2^{\gamma}$ more such subsets, i.e., at least $2^{\gamma}$ more bicliques that can be found in step 2.  Thus in this case there exist enough bicliques that step 2 can find $k$ of them.


In either case, if the graph has at least $k$ bicliques, this algorithm will output at least $k$ of them.