$\DeclareMathOperator\BB{BB}$My answer is a slight cheat, because I'm going to give two examples where we have not only not discovered any algorithms but where we know there is no such algorithm. Both use the Halting theorem, although the first more obviously than the second.

The first is the Busy Beaver number: For any given $n$, there are finitely many Turing machines with at most $n$ states which halt when run on the blank tape. (Here to deal with ambiguity issues we assume that are only symbols are 1 and 0). Of those machines which halt, there is some maximum time any of them goes before halting. What is it? This is a function of $n$, sometimes denoted $\BB(n)$. It is clear from the definition that this quantity is finite. However, finding $\BB(n)$ is equivalent to solving the Halting problem. For a good introduction to the Busy Beaver function an related work, see Scott Aaronson's survey here.

Since the last one involved Turing machines, it probably isn't surprising that it has no algorithm. The second has a similar flavor, but where it is less obvious from the statement: For any given $n$, there are only finitely many Diophantine equations with at most $n$ variables, degree at most $n$, and no coefficient greater than $n$. Of those equations, some have no solutions. Define $D(n)$ to be the number of those equations which have no solutions. As before $D(n)$ is clearly finite. In this case, the problem is again equivalent to the Halting problem. In this case, it is due to the solution of Hilbert's tenth problem, which shows this problem to be equivalent to the Halting problem.

Two tangential remarks: First, if one defines a function similar to $D(n)$, $F(n)$ which instead counts the largest number of solutions of any Diophantine equation with at most $n$ variables, degree at most $n$, and no coefficient greater than $n$, then this number is also finite. But I do not know a proof or disproof that there is an algorithm for $F(n)$. Second, I don't know of any examples similar to these where we can prove finiteness and know there is no algorithm, but where the proof doesn't essentially go through the Halting theorem.

one single knownalgorithm which outputs the set of counterexamples to smooth Poincaré.) $\endgroup$