What is the remainder ?

Writing the given equation $3m=4^p-1$ modulo $p$ , we find $3m\equiv4-1={3}$ by Fermat's little theorem, so that $m\equiv{1} \pmod{p}$ . Since $m-1$ is even, we can write $m-1=2pn$ for some $n$ . Now $2^{m-1}=2^{2pn}=(4^p)^n=(3m+1)^n\equiv{\boxed{1}} \pmod{m}$

Substituting $p = 2$ to the left equation, then $m = \frac { 4 ^ { 2 } - 1 } { 3 } = \frac { 16 - 1 } { 3 } = \frac { 15 } { 3 } = \boxed { 5 }$ .

Substituting $m = 5$ to the right expression, we get $2 ^ { 5 - 1 } \mod 5 = 2 ^ { 4 } \mod 5 = 16 \mod 5 = \boxed { 1 }$ .

So, the answer is $\boxed { 1 }$ .