Is this attack for RSA possible?
The original question states that we are given $e,m_1,{m_1}^e,{m_1}^d,\dots,{m_x}^e,{m_x}^d$, but not $N$; and asks for $d$ or a factorization of $N$.

I suspect some gremlin crept in this statement, because if any of the ${m_j}$ is such that ${m_j}&gt;1$, it is trivial to obtain $d$ from ${m_j}$ and ${m_j}^d$, as $d=\log({m_j}^d)/\log(m_j)$.

So I guess we are given $e, m_1, {m_1}^e\bmod N,{m_1}^d\bmod N,\dots,{m_x},{m_x}^e\bmod N,{m_x}^d\bmod N$. If we are not given $N$, and for non-trivial choice of messages, we can find $N$ with overwhelming probability as $N=\gcd_{j=1}^x(({m_j}^e)-({m_j}^e\bmod N))$. With $N$ known, the givens ${m_1}^e\bmod N$ can be eliminated, for we can recompute them.



Thus the problems likely boils down to: given $N,e,m_1,{m_1}^d\bmod N,\dots,m_x,{m_x}^d\bmod N$, can we find $d$ or a factorization of $N$?

If the givens (beside $N$) allowed to efficiently factor $N$, that would solve Open Problem 1 in Dan Boneh's survey given as the question's second reference, which is nearly that problem (in addition, there is freedom to choose the $m_j$).

Thus no, we do not know how the givens (beside $N$, or the givens that let us find $N$) help in factoring $N$, but we do not know how to prove that they do not help significantly. That's an open problem!

We however can state one provable fact: if the $m_j$ are chosen at random, then knowledge of the ${m_j}^d\bmod N$ can not help. Proof sketch: any efficient algorithm factoring $N$ from $(N,e,m_1,{m_1}^d\bmod N,\dots,m_x,{m_x}^d\bmod N)$ with the $m_j$ chosen at random could be turned into an efficient algorithm that factors $N$ from $(N,e)$ alone, by choosing some $c_1,\dots,c_x$ at random and computing the $m_j={c_j}^e\bmod N$.



Some triva:

If we knew $d$, or any working private exponent $d'$, then we could factor $N$ using an efficient probabilistic algorithm.

If we knew the factorization of $N$, we could find the smallest working private exponent $d_0=e^{-1}\bmod\lambda(N)$, such that $d\equiv d_0\pmod{\lambda(n)}$. And if we further knew that $d=e^{-1}\bmod\phi(N)$ (rather than just some working private exponent), we could find $d$.