Serbian Mathematical Olympiad

Solution:

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
$$\left(2^{2015}+1\right)^{x}+2^{2015} \equiv 2^{2015} \equiv 5 \pmod{9},$$
hence $2^{y} \equiv 4 \pmod{9}$, from which we obtain $y=6k+2$ 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 $2^{2015} \equiv 7 \pmod{13}$, so we obtain $8^{x}+7 \equiv \pm 4 \pmod{13}$. However, $8^{x}$ gives one of the residues $1,5,8,12$ modulo $13$, so the last congruence is impossible.

Solution:

For $x>1$ we have
$$2^{y}=\left(2^{2015}+1\right)^{x}+2^{2015}-1=(x+1)2^{2015}+\sum_{i=2}^{x}\binom{x}{i}2^{2015i} \equiv (x+1)2^{2015} \pmod{2^{2019}},$$
so since $y>2019$ it follows that $16 \mid x+1$.

Now consider the equation modulo $17$. Since $2^{2015} \equiv 9 \pmod{17}$, we obtain $2^{y} \equiv 10^{x}+8 \equiv 10^{15}+8 \equiv 3 \pmod{17}$, which is impossible.