Assuming x, y, z are non-negative integers, there are no solutions to $4^x + 7^y = z^2$ and $4^x + 11^y = z^2$. This can be shown through modular arithmetic.
Taking the equation a) mod 3 you'll find the following:
- $4^x \equiv 1 \pmod{3}$
- $7^y \equiv 1 \pmod{3}$
- $z^2 \equiv 0$ or $1 \pmod{3}$
Since $1 + 1 = 2$ and there are no squares equivalent to $2$ (mod 3), there are no solutions to $4^x + 7^y = z^2$.
Now taking equation b) mod 3:
- $4^x \equiv 1 \pmod{3}$
- $11^y \equiv (-1)^y \pmod{3}$
- $z^2 \equiv 0$ or $1 \pmod{3}$
Since, once again, there are no squares equivalent to $2$ (mod 3), $y$ must be odd. However taking the equation mod 4 you'll get the following:
- $4^x \equiv 0 \pmod{4}$
- $11^y \equiv (-1)^y \pmod{4}$
- $z^2 = 0$ or $1 \pmod{4}$
$y$ cannot be odd since this would imply that $11^y$ is equivalent to $-1 \equiv 3$ (mod 4) and there are no squares equivalent to 3 mod 4. This is a contradiction and therefore there are no solutions to equation b) either.
EDIT: Actually I just realised that I neglected the case where $x = 0$ and thus $4^0 \equiv 1 \pmod{4}$. But taking the resulting equation $1 + 11^y = z^2$ mod 5 proves that there's no solution anyway, as the quadratic residues for modulo 5 are 0, 1 and 4.