6. In the set of non-negative integers, solve the equation

$$
\left(2^{2015}+1\right)^{x}+2^{2015}=2^{y}+1 . \quad \text { (Bojan Bajic) }
$$
6. For $x \leqslant 1$, the only solutions are $(0,2015)$ and $(1,2016)$.

Let $x>1$. Since $2^{2015}+1$ is divisible by 3, we have $\left(2^{2015}+1\right)^{x}+2^{2015} \equiv 2^{2015} \equiv 5$ $(\bmod 9)$, hence $2^{y} \equiv 4(\bmod 9)$, from which we get $y=6k+2$ for some $k \in \mathbb{N}$. Now modulo 13, we have $2^{y}+1=\left(2^{6}\right)^{k} \cdot 2^{2} \equiv \pm 4$ and $2^{2015} \equiv 7(\bmod 13)$, so we get $8^{x}+7 \equiv \pm 4(\bmod 13)$. However, $8^{x}$ gives one of the remainders $1,5,8,12$ modulo 13, so the last congruence is impossible.

Second solution. For $x>1$ we have $2^{y}=\left(2^{2015}+1\right)^{x}+2^{2015}-1=(x+1) 2^{2015}+$ $\sum_{i=2}^{x}\binom{x}{i} 2^{2015 i} \equiv(x+1) 2^{2015}\left(\bmod 2^{2019}\right)$, so because $y>2019$ it follows that $16 \mid x+1$. Now consider the equation modulo 17. Since $2^{2015} \equiv 9(\bmod 17)$, we get $2^{y} \equiv 10^{x}+8 \equiv 10^{15}+8 \equiv 3(\bmod 17)$, which is impossible.