<s>[INST] Solve for all $x,y\in N$ and $x^3-3^y = 2015$ Question:

Solve for all $x,y\in N$ and $x^3-3^y = 2015$

So far I have tried $\mod(7)$ and $\mod(8)$ but I am not sure if I am in the right path or not I will be happy if someone can toss me a hint.
 [/INST] Use Metin's idea of "reject solution of too large" and
someone's idea of mod $13$ -> $3\mid m$, I reach a solution.
Consider mod $13$, then
$3^y\equiv1,3,9 \pmod {13}$
$x^3\equiv1,5,8,12 \pmod {13}$
Thus, $3^y\equiv1 \pmod {13}, x^3\equiv1 \pmod {13}, 3\mid y$
Let $y=3z, k=3^z$, so $x>k>0$
Then $x^3-3^y = x^3-(3^z)^3 = x^3-k^3 = (x-k)(x^2+xk+k^2) = 2015 $
$x^2+xk+k^2$ is factor of $2015$ and thus bounded above, we can have $x<45$. Then the rest is check and reject by computer. (By hand is tough)
Furthermore, consider $x^3-27^z=2015$ with mod $7$, we have $7\mid x$. (Consider all outcomes as above). With $x\equiv2 \pmod{6}$ by Alexander. We can reject all $x<45$ except $14$.
</s>