Solve a diophantine equation The questions is: Solve in the positive integers $$3^n=m^4+m+1.$$
I can prove that $n$ is odd, and $m\equiv 4(\text{mod }9)$, but i dont know why $n$ and $m$ are $1$.

Probably not of much use but some elementary observations about the numbers $m^4+m+1$ are as follows.
There is precisely one solution modulo $3^n$ of the equation 
$$m^4+m+1\equiv 0\pmod {3^n}$$
The solution for $3^{n+1}$ can be obtained from the solution for $3^n$ as follows:- 
If $m^4+m+1\equiv 3^nd\pmod {3^{n+1}}$, where $d=1$ or $2$, then 
$$(m+3^nd)^4+(m+3^nd)+1\equiv 0\pmod {3^{n+1}}$$
The first few solutions are then
$m=1:$ $m^4+m+1\equiv 3\pmod {3^2}$
$m=1+3=4:$ $m^4+m+1\equiv 2\times 3^2\pmod {3^3}$
$m=4+2\times 3^2=22:$ $m^4+m+1\equiv  3^3\pmod {3^4}$ 
$m=22+3^3=49:$ $m^4+m+1\equiv 2\times 3^4\pmod {3^5}$
$m=49+2\times3^4=211:$ $m^4+m+1\equiv 3^7\pmod {3^8}$ 
$m=211+3^7=2398:$ $m^4+m+1\equiv 2\times 3^8\pmod {3^9}$ 
