Checking acceptance of a word vs finding an accepted word
The two problems are Turing equivalent: one is undecidable for Turing machines iff the other is.

Deciding whether a Turing machine (TM) $M$ halts on a specific input $x$ is the Halting problem for Turing machines.

Given a TM $M$, we want to know whether it halts on some word. From
$M$ we build a TM $M^*$ that, independently of its input, dovetails the computation of $M$ over all
inputs (i.e. it mimics the computation of $M$ on all possible inputs in
parallel), and halts iff one of these computations halts.
If we had an oracle $H$ that decided the halting problem, it could
decide whether $M*$ halts or not, hence whether the TM $M$ halts on
some input. This is only for the decision on the existence of that
input, but the same dovetailing construction could then be used to
compute the halting input when it is known to exist. Hence we could build a TM that always halts, either with a word on which $M$ halts if there is one, or with failure when there is none.

Hence, if the halting problem were decidable, the problem of deciding
whether a TM halts on some input, and determining that input wold be
decidable too.

Conversely, let us assume that it is decidable whether a TM halt on
some input. Then given a Turing machine $M$ and and input $x$ we want
to decide whether $M(x)$ halts (the halting problem for $M$ on input
$x$).  For this we build a machine $M'$ that, independently of its
input, will simply mimic the machine $M$ running on input $x$. Hence
the machine $M'$ halts on some input (thus on any input) iff $M$ halts
on input $x$. Hence we have a decision procedure for the halting
problem if we have a decision procedure for halting on some input.

Both problems are Turing reducible to each other. Hence they are
Turing equivalent. If one were to be decidable, the other would be too. Or, to take the contrapositive, the undecidability of one implies the undecidability of the other.