We claim that the answer is all $ m\ne 1$. Clearly, $ m = 1$ yields $ n - 1$, which can clearly be a mutliple of any prime. Now, we consider $ m > 1$. Let $ p$ be an arbitrary prime factor of $ m$. Let $ p^k || m$, so let $ p^k l = m$, where $ \gcd (l, p) = 1$. Now, assume that no prime exists so that $ n^m - m \equiv 0\bmod q$ has no solution for $ n$. Set $ (p^kl)^{p - 1} + (p^kl)^{p - 2} + \cdots + p^kl + 1 \equiv 0 \bmod q$. Since the Left Hand Side of this is not congruent to $ 1\bmod p^{k + 1}$, we may set $ q$ not congruent to $ 1\bmod p^{k + 1}$. We will show that this $ q$ yields a contradiction. First, note that the remainder of $ m^{p - 1} + m^{p - 2} + \cdots + 1$ when divided by $ m - 1$ is $ 1 \cdot p = p$, which is relatively prime with $ m - 1 = p^kl - 1$. Thus, $ \gcd (q, m - 1) = 0$, so $ m$ is not congruent to $ 1\bmod q$. Since $ n^m \equiv m \bmod q$, we have that $ n^{p^kl} \equiv p^kl\bmod q$, so $ n^{p^{k + 1}l}\equiv (p^kl)^p\equiv 1\bmod q$. Let the order of $ n\bmod q$ be $ x$. This means that $ x|(p^{k + 1}l)$. However, since $ n^{p^kl}\equiv p^kl\bmod q$, which is not congruent to $ 1\bmod q$, we have that $ x$ is not a factor of $ p^kl$, so $ p^{k + 1}|x|(q - 1)$, so $ q\equiv 1\bmod p^{k + 1}$, which is a contradiction. Thus, there exists a prime, $ q$, so that $ q$ is not a factor of $ n^m - m$ for all integers $ n$.

$ \gcd (q, m - 1) = 0$, so $ m$ is not congruent to $ 1\bmod q$. ... ... However, since $ n^{p^kl}\equiv p^kl\bmod q$, which is not congruent to $ 1\bmod q$, we have that $ x$ is not a factor of $ p^kl$, so $ p^{k + 1}|x|(q - 1)$, so $ q\equiv 1\bmod p^{k + 1}$, which is a contradiction.