Recently, Miyazaki and Togbé proved that for any fixed odd integer b ≥ 5 with b ≠ 89, the Diophantine equation $b^x+2^y={(b+2)}^z$ has only the solution (x,y,z) = (1,1,1). We give an extension of this result.

It is shown that for a given squarefree positive integer D, the equation of the title has no solutions in integers x > 0, m > 0, n ≥ 3 and y odd, nor unless D ≡ 14 (mod 16) in integers x > 0, m = 0, n ≥ 3, y > 0, provided in each case that n does not divide the class number of the imaginary quadratic field containing √(-2D), except for a small number of (stated) exceptions.

Let $\mathbb{Z}$, $\mathbb{N}$ be the sets of all integers and positive integers, respectively. Let $p$ be a fixed odd prime. Recently, there have been many papers concerned with solutions $(x,y,n,a,b)$ of the equation $x^2+2^a{p}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$, $b\ge 0.$ And all solutions of it have been determined for the cases $p=3$, $p=5$, $p=11$ and $p=13$. In this paper, we mainly concentrate on the case $p=3$, and using certain recent results on exponential diophantine equations including the famous Catalan equation, all solutions $(x,y,n,a,b)$ of the equation $x^2+2^a·{17}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$, $b\ge 0$, are determined....

We consider $N$, the number of solutions $(x,y,u,v)$ to the equation ${(-1)}^{u}r{a}^x+{(-1)}^{v}s{b}^y=c$ in nonnegative integers $x,y$ and integers $u,v\in \{0,1\}$, for given integers $a\>1$, $b\>1$, $c\>0$, $r\>0$ and $s\>0$. When $gcd(ra,sb)=1$, we show that $N\le 3$ except for a finite number of cases all of which satisfy $max(a,b,r,s,x,y)\<2·{10}^{15}$ for each solution; when $gcd(a,b)\>1$, we show that $N\le 3$ except for three infinite families of exceptional cases. We find several different ways to generate an infinite number of cases giving $N=3$ solutions.