Why we can't use non-deterministic turing machines in this case?

Can you explain me why we can't show decidability of problem using non-determinism ? I know that this problem (described below) is not decidability, however I can't understand why following reasoning is not working.

Problem
Given a context free grammar $G$, does there exists a word $w∈Σ^∗$ such that $w^4∈L(G)$ ?

My reasoning:
Lets show Turing machine which decide this problem. Let machine guesses some word $w$ (it is possible because word is finite and $Σ$ is finite) and check if (using CYK algorithm) $G$ generates $w^4$.

I totally don't understand why this approach doesn't work. After all, non-determinism is about guessing...

Your reasoning is not wrong. But recall that decidability requires TM machine halt with YES if $\exists w \text{ such that } w^4 \in L(G)$ or NO if $\nexists w \text{ such that }w^4 \in L(G)$. In other words, the TM always halts.

In your approach the TM machine has to guess infinitely many strings and check if their forth power belongs to $L(G)$. A nondeterministic Turing machine cannot check infinitely many strings in finite amount of time.

Decidability problems efficiently solved by non-deterministic Turing machines are also solved by deterministic Turing machines. In other words, their computational powers are equivalent.

StackExchange Computer Science Q#79678, answer score: 6