Just knowing any such triple would not help at all. Reason:
Choose $g$ and $x$ arbitrarily, calculate $h$. Now you got such a triple, but it shouldn't help you factorize $N$, unless you have another interesting property.
If on the other hand you can somehow calculate roots (given $x$ and $h$, and then find $g$), then you can break RSA: Choose $x=e$, and your root algorithm finds the plaintext to a ciphertext.
If you can find square roots (and there are always 4 of them with $n=pq$), then you can actually factorize $N$, the attack basically works like the chosen ciphertext attack on Rabins encryption scheme: Choose $a$, calculate $a^2$, apply your algorithm to $a^2$ to find a square root. Since it only has $a^2$ as input, it will result in one of the four square roots, but not necessarily the one you started with. If the result is not $a$ or $-a$, then you can factorize $N$.