Solve : $\space 3^x + 5^y = 7^z + 11^w$ 
Solve the diophantine equation $3^x + 5^y = 7^z + 11^w$,here $x,y,z,w$ are all non-negative integers. 

I find three solutions by force algorithm use Mathematica: (0,0,0,0)(1,1,1,0)(1,3,1,2).And there is no else when $y,z,w<100$.
Thanks in advance.  
 A: This is just a partial answer/analysis because I feel it is not fitting as a comment.
Edit: Ivan Loh's excellent answer gives a full resolution.

Let us employ modulo arithmetic to reduce some cases:
Modulo $3$: This assumes $x\ne 0$: $0^x+(-1)^y \equiv 1^z +(-1)^w \pmod 3$, hence $y$ odd, $w$ even. 
Modulo $4$: $(-1)^x + 1 \equiv (-1)^z + 1 \pmod 4$ (since $w$ even), hence $x,z$ have equal parity.
Modulo $5$: This assumes $y \ne 0$: $3^x \equiv 2^z +1 \pmod 5$. We can enumerate all solutions by letting $x$ run. For $x=0$ there is no solution; $x=1, z=1$; $x=2,z=3$; $x=3,z=4$. Because $\Bbb Z_5^\times$ has order $4$, we conclude that $x=z=1 \pmod 4$.
Modulo $7$: This assumes $z \ne 0$: An exhaustive search reveals that modulo $6$, we have the following six options for $(x,y,w)$:
$$(1,1,0) \quad (1,3,2) \quad (3,1,4) \quad (3,5,2) \quad (5,3,4)\quad (5,5,0)$$

In particular, we have $z \ne 0$ implies $y \ne 0$, which in turn implies $x \ne 0$, which by the modulo 4 case is equivalent to $z \ne 0$. Thus if $x = 0$, also $y=z=0$ and we miss only the trivial solution $(x,y,z,w)=(0,0,0,0)$ by assuming $x,y,z$ nonzero.
