We know that $5a^b=2004+b$ then $b=5c+1$ for some integer $c$. Then $a^{5c+1}=401+c$. If $a>2$ then $c=0$. Hence $a=401$ If $a=2$ then $2^{5c+1}=401+c$ and there are no such solution. Hence the solution is $a=401,b=1$

How do we know that if $a > 2$, $c = 0$?

KKMathMan wrote: How do we know that if $a > 2$, $c = 0$? If $a>2, c\neq 0$, then $$a^{5c+1}\geq 3^6=729>401+3$$If any such $a$ is fixed then, $LHS>RHS$ and if any such $c$ were fixed, then still $LHS>RHS$. Thus, the given two cases are valid.