<s>[INST] Solve the following in non-negative integers: $3^x-y^3=1$. Solve the following in non-negative integers: $$3^x-y^3=1$$
Of course $(x,y)=(0,0)$ is a trivial solution. After seeing that I proceeded like this:
$$3^x-y^3=1$$$$\implies3^x-1=y^3$$$$\implies2(3^{x-1}+3^{x-2}+ \cdots +3^1+1)=y^3$$$$\therefore2|y$$
So let $y=2k$ for non-negative $k$.
Then $$3^{x-1}+3^{x-2}+ \cdots +3^1+1=4y^3$$ So $4|LHS$ using a mod argument one can easily deduce that this implies that $x$ is even so let $x=2q$.
Then going back to our original expression we have $$9^q-1=8k^3$$$$\implies 9^{q-1}+9^{q-2}+ \cdots +9+1=k^3$$
Then again using mod arguments I deduced that $k \equiv 4 \pmod9$. And after that things started to get more and more yucky. This could be a completely bad approach so I also triad looking at $(y+1)(y^2-y+1)=3^x$ but that didn't lead me anywhere either. 
Thanks in advance for any help.
 [/INST] If $y+1=3^a$ with $a\ge1\iff y>1$
$y^2-y+1=3(3^{2a-1}-3^a+1)$ which can not be power of $3$ unless $2a-1=a\iff a=1$
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