Clearly $m$ and $n$ are both odd; and moreover they are coprime. Now that $(m,n)=(1,1)$ works, let us show there exists no other solutions. For this, we inspect the number $2m^5-2n^5-1$. Note that $2m^5-2n^5-1\equiv 2m^5-1\equiv 0\pmod{n^4}$ and moreover, $2m^5-2n^5-1\equiv -(2n^5+1)\equiv 0\pmod{m^4}$. Since $m,n$ are coprime, it follows that \[ m^4 n^4 \mid 2m^5 - 2n^5-1. \]Now note that $2m^5-2n^5-1\ne 0$ due to parity considerations. This leaves us with two cases to inspect. Case 1. Assume $2m^5>2n^5+1$. We then have $m^4n^4 +2n^5+1 \le 2m^5$. Note also that from the second divisibility, it follows $n^5 \ge \frac{m^4-1}{2}$. Hence, $n^4 \ge m^2$ (this is not hard to see - otherwise if $n^4<m^2$ then $n<\sqrt{m}$ which gives a contradiction). But then $m^4n^4 + 2n^5+1 \ge m^6+2n^5+1 \ge 3m^5+2n^5+1$, which is clearly greater than $2m^5$. Case 2. Now suppose $2n^5+1>2m^5$. This yields $m^4n^4 + 2m^5\le 2n^5+1$. Now, $m^5\ge \frac{n^4+1}{2}$, hence $m^4>2n$ (easy to show). This will yield the desired contradiction.