First we suppose that $y \geq 2$. -$1+2^y=(-1)^z \mod 4$ thus for $y \geq 2$ we have $z$ is even. -By looking modulo $5$ we get $3^x+(-1)^y=0$ if $x=2a+1$ for some $a$ then $3^x+(-1)^y=3(-1)^a+(-1)^y=0 \mod 5$ but this is impossible. Hence $x$ is even. Let $x=2a$ and $z=2b$ and we suppose that $a,b \geq 1$ then the equation becomes : $$(2015^b-2013^a)(2015^b+2013^a)=2014^y$$ Let $d=\gcd(2015^b-2013^a,2015^b+2013^a)$ then $d \mid 2.2015^b$ and $d \mid 2.2013^a$ thus $d \mid 2 \gcd(2013^a,2015^b)=2$ and both $2015^b-2013^a$ and $2015^b+2013^a$ are even.Therefore $d=2$ Now $\frac{2015^b-2013^a}{2}\frac{2015^b+2013^a}{2}=1007^y2^{y-2}$ and the terms in the product of the left hand side are coprimes. Since $1007^y>2^{y-2}$ and $\frac{2015^b-2013^a}{2}<\frac{2015^b+2013^a}{2}$ then : $$\mbox{ either } \frac{2015^b-2013^a}{2}=1, \frac{2015^b+2013^a}{2}=1007^y2^{y-2} \mbox{ or } \frac{2015^b-2013^a}{2}=2^{y-2}, \frac{2015^b+2013^a}{2}=1007^y$$ -In the first case $ 2015^b=2013^a+2$ so if $a \geq 2$ then modulo $9$ we have $ (-1)^b=2 \mod [9]$ which it is a contradiction thus $a \leq 1$. $a=1$ implies $b=1$ and $x=z=2$ so $2014^y=2015^2-2013^2=4.2014$ so there is no solution in this case. $a=0$ leads to $2015^b=3$ which it is impossible. -In the second case if $a \geq 2$. Then $2013^a=1007^y-2^{y-1}=(-1)^y-2^{y-1} \mod [9]$ but this equation doesn't have a solution. Hence $a=0$ or $a=1$. If $a=0$ then $1+2^{y-1}=1007^y>2^y$ is impossible. If $a=1$ then $2013+2^{y-1}=1007^y>2^{9y}$ so $2013 >2^{y-1}(2^{8y+1}-1)>2(2^{8y+1}-1)$ so $y \leq 1$. Now we study $y=0$ and $y=1$ : $y=0$ implies $2013^x+1$ is even which it is impossible with $2015^z$ to be odd. $y=1$ implies $2013^x+2014=2015^z$ so if $x \geq 2$ then $9 \mid 2015^z+2$ this is impossible because $2015 = -1 \mod [9]$. Hence $x\leq 1$. $x=1$ gives us $2015^z=4027$ which cannot be solved in the set of integers. $x=0$ leads to $z=1$ . Finally the only solution is $(0,1,1)$.

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