Does RSA have two trapdoors?
Defining RSA Problem as finding $m$, having $N,e,c$ where $c=m^e \mod N$ and $N=pq$, two unknown same-size prime numbers.

So far the best generic method we know to solve this is to factor $N$.

But there might be other methods, so far unknown, to compute the $e$-th root of $c$ and recovering $m$.

This is different from the discrete logarithm problem, where the exponent is secret. This is just computing an $e$-th root. We know how to compute roots modulo a prime, we don't know how to compute roots modulo a composite number of unknown factorization.

Note: for bad choices of m (i.e. without the use of padding) or $e$ or $d$ (the private exponent) there are practical attacks that recovers $m$ without factoring. For example the Weiner's or Coppersmith's attack. In particular the method of Coppersmith does compute $e$-th roots if $e$ and $m$ are small enough therefore it is not a generic method.

The RSA Problem is at least as easy as factoring. But there might be other, so far unknown, easier way to compute $e$-th roots without factoring.