Yes, every undecidable (not semi-decidable) language has this property.
For example, consider the set $L = \{(x,M) \mid M \text{ does not halt on input } x \}$.
Suppose we have an algorithm that can enumerate the members of this set. If such an algorithm existed, we could use this to solve the halting problem with inputs $x,M$, with the following algorithm:
- Alternate between running machine $M$ for $n$ steps on $x$, and enumerating the $n$th member of $L$.
$M$ either halts, or does not halt on $x$. If it halts, eventually we will find an $n$ where we reach a halting state. If it doesn't halt, then eventually we will reach $(M,x)$ in our enumeration.
Thus we have a reduction, and we can conclude that no such enumeration exists.
Note that such enumerations can exist for semi-decidable problems. For example, you can enumerate the set of all halting machine-input pairs by enumerating all possible traces of all Turing Machine executions after $n$ steps, and filter out ones that do not end in a halting state.