Integer solutions to $x^{x-2}=y^{x-1}$ 
Find all $x,y \in \mathbb{Z}^+ $ such that $$x^{x-2}=y^{x-1}.$$

I can only find the following solutions: $x=1,2$. Are there any other solutions? 

The only solutions are $x=1$ and $y=$ anything, and $x = 2$ and $y = 1$.
Start by rearranging the equation: $(\frac{x}{y})^{x-1} = x$.
If x = 1 then we have: $(\frac{1}{y})^0 = 1$ which is true for any $y$.
If y = 1 then we have: $x^{x-1} = x \implies x^{x-2} = 1$, which means either $x = 1$ which is covered by the case above, or $x = 2$.
If $x \not = 1$, $y \not = 1$ and $x$ is not a multiple of $y$, then $(\frac{x}{y})^{x-1} = x$ is in the form $\text{fraction}^{\text{positive integer}} = \text{positive integer}$, which is always false.
If $x \not = 1$, $y \not = 1$ and $x$ is a multiple of $y$, then we can write $x = ky$ for some $k \in \mathbb{Z}^+$. So we get $k^{ky-1} = ky \implies k^{ky-2} = y \implies (k^k)^y = k^2y$. If $k = 1$ then $y = 1$ and $x = 1$ (covered), or if $k = 2$ then $y = 1$ and $x = 2$ (covered), so any other solutions has $k > 2$. We see then that at $y = 1$, the $\text{LHS} > \text{RHS}$. You now want to show that it's true for every $y$. You can do that by doing the following:
Let $u_y = (k^k)^y$,
$v_y = k^2y$. Compute $\frac{u_y}{u_{y-1}} = k^k \geq 27$, $\frac{v_y}{v_{y-1}} = \frac{y}{y-1} = 1 + \frac{1}{y-1} \leq 2$. So $\frac{u_y}{u_{y-1}} > \frac{v_y}{v_{y-1}}$. So $u_y > v_y, \forall y$. No more solutions.
