[This answer has been extensively edited to address the question more fully and correct minor errors.]
Diophantine equations are often very difficult. However, these two equations yield to fairly simple methods.
Since $z^2 > 0$, it is required that $(x^y – y^x) / (x – y) > 0$. To avoid division by zero, we must have $x \ne y$. Suppose that 0 < x < y.
Case 1: $x \geq 3$. Then $x^y > y^x$, and $(x^y – y^x) / (x – y) < 0$, so there are no solutions.
Case 2: x = 2. (x, y, z) = (2, 3, 1) is a solution, as stated in the question. If $y \geq 4$, then $2^y \geq y^2$ so there are no more solutions.
Case 3: x = 1. Then $(x^y – y^x) / (x – y) = (1 – y) / (1 – y) = 1$. Hence (x, y, z) = (1, n, 1) for any positive integer $n \geq 2$. This shows that the number of solutions is infinite. Note however that, as cases 2 and 3 illustrate, an infinite set of solutions to an equation does not necessarily contain all solutions.
For 0 < y < x, simply reverse the values of x and y in the above solutions (since $(x^y – y^x) / (x – y)$ is symmetrical in x and y).
A formula which gives all the solutions with x < y is (x, y, z) = (1+a, 1+b(1-a)+2a, 1) where a = 0 or 1 and b is any positive integer. However, this is a bit artificial.
This can be rewritten as:
$[x^3 - x] + [y^2 - y] + [x^2y^3 - x^2y - xy^2] = 0$
Then if x, y $\geq$ 2, each of the three terms enclosed by square brackets must be positive. To see this for the third term, note that if x, y $\geq$ 2:
$x^2y^3 \geq 2x^2y^2 > (x^2y + xy^2)$
Hence there are no solutions with $x, y \geq 2$. The only remaining possibilities for a solution in positive integers are x = 1 or y = 1. If x = 1 then:
$1 + y^2 - y + y^3 - y^2 = 1 + y$
$-y + y^3 = y$
$y(y^2 - 2) = 0$
Hence y = 0 or +/-$\surd2$ and so is not a positive integer. A similar argument shows that if y = 1 then x is not a positive integer. Hence equation 2 has no solution in positive integers (and questions 2-4 are not relevant to it).
Generalising from these problems
One lesson is to try simple approaches first. In the case of equation 1 the variables as exponents may suggest that it is very difficult. But if one asks simple questions the apparent difficulties can - in some cases - disappear. Key questions here are: Under what conditions is $(x^y - y^x)/(x-y) > 0$? What is the effect of putting x = 1? But there are very many diophantine equations for which such simple methods will not work.