Solve in positive integers the equation $3^x+4^y=5^z$.
My thoughts: I see that $x=2, y=2, y=2$ is a solution set from Pythagorean theorem and $x=0, y=1, z=1$ is a solution set from just plugging in numbers, but is there a method to solve these for solution sets and/or show these solutions are unique?
Note that:
$$3^x+4^y=5^z\tag{1}$$
translates into:
$$1^y\equiv(-1)^z\quad\text{(mod 3)}$$
This means that $z$ has to be even or:
$$z=2z_1$$
With that in mind the equation (1) becomes:
$$3^x+4^y=5^{2z_1}$$
$$3^x+2^{2y}=5^{2z_1}$$
$$3^x = 5^{2z_1}-2^{2y}$$
$$3^x = (5^{z_1}-2^{y})(5^{z_1}+2^{y})$$
Notice that $5^{z_1}$ and $2^{y}$ are not divisible by 3 and their sum and difference cannot be both divisible by 3 at the same time (the proof is elementary). So one factor (smaller) must be 1 and the second factor (bigger) must be a power of 3:
$$5^{z_1}-2^y=1\tag{2}$$ $$5^{z_1}+2^y=3^x$$
According to Catalan's conjecture (or Mihăilescu's theorem) equation (2) has no solutions for $y,z_1>1$. What remains to be done is to test a trivial case when $y=1$ or $z_1=1$.
For $y=1$ there is no solution for $z_1$ but for $z_1=1$ (which means $z=2$) we have a solution $y=2$.
If you replace $y=2$ and $z=2$ into (1), you get $x=2$.
There are no other solutions.
EDIT:
You don't have to use the Catalan's conjecture to prove that equation:
$$1 + 2^a=5^b\tag{3}$$
...has only one solution ($a=2,b=1$). Suppose now that $a\ge3$.
It means that LHS of (3) is equal to 1 (modulo 8). On the other side (again modulo 8): $5^1\equiv5$, $5^2\equiv1$, $5^3\equiv5$, $5^4\equiv1$...
So $b$ must be even, or $b=2b_1$.
Now consider the same equation, this time modulo 3:
$$1 + (-1)^a\equiv(-1)^{2b_1}$$
$$1 + (-1)^a\equiv1$$
...which is impossible for any value of $a$.