Solutions of some Diophantine equations Respected Mathematicians,
The Diophantine equation

$$2^x + 5^y = z^2$$

has solutions $$x = 3, y = 0, z = 3$$ and $$x = 2, y = 1, z = 3$$ I got these solutions by trial and error method. To be honest, these solutions are below the number $5$. So, I easily verified them by trial and error method. I would like to know the method which will give the solutions of the above equation, as well as the solutions of equations below.
a)

$$4^x + 7^y = z^2$$

b)

$$4^x + 11^y = z^2$$

Looking forward to your solution and support.
baba

I'll do a piece of it to show you some methods you can try on the other pieces. 
$2^x+5^y=z^2$. Let's do the case where $y=2s$ is even. 
$2^x=z^2-(5^s)^2=(z+5^s)(z-5^s)$, so $z+5^s=2^m$ and $z-5^s=2^n$ with $m+n=x$. Eliminating $z$, $2\times5^s=2^m-2^n$, so $5^s=2^{m-1}-2^{n-1}$. The left side is odd, so the right side is odd, so $n=1$, and $5^s=2^{m-1}-1$. Left side is 1 modulo 4, so right side is 1 modulo 4, so we must have $m=2$. So if there's a solution with $y$ even, then $x=3$, $y=0$, $z=3$. 
