Note: klugreuter's answer outlined a different approach to the problem, that I developed here. This obsoletes the following, except perhaps when we want a small $e$.
We can compute $\mathbin|m^e-c\mathbin|$ for various small odd $e>1$, and try to factor them (even partially). As soon as we get two distinct primes factors $p$ and $q$ for some $\mathbin|m^e-c\mathbin|$, with $\gcd(p-1,e)=1$ and $\gcd(q-1,e)=1$, $m<p\,q$ (and $c<p\,q$ if that's added to the problem statement), then we can compute $N$ and a $d$ matching $e$ for this $N$ by one of the methods used in RSA, and that solves the problem. That's at least sometime feasible by pulling moderate factors using ECM factoring (e.g. using GMP-ECM).
If $N$ is required to be large (including, because $m$ is; $c$ has lesser influence), it is hard to find $p$ and $q$ with large enough a product. But we can sometime find more than two distinct large primes dividing $m^e-c$, and compute $N$ and $d$ as in multi-prime RSA; that improves our chances to find $(N,e)$ passing scrutiny (but does not answer the problem as worded, since it requires $N$ to be a bi-prime).
Addition: in the question as asked, there's no requirement that $c<N$ (that's normally part of RSA). When $m\ll c$, that allows to try the above with $e=1$, that is factor $c-m$, which is much smaller than above. If we get distinct prime factors $p$ and $q$ with $m<p\,q$ that gives a solution. We can hide that we started from $e=1$ by adding a multiple of $\operatorname{lcm}(p-1,q-1)$ to $e$ and/or $d$.
Late addition/hint: if we are allowed to use $e=\operatorname{lcm}(p-1,q-1)+1$ as hypothesized above, then at least one choice of $e$ not discussed above is worth consideration.