How do I solve the system $x^{x+y} = y^4$ and $y^{x + y} = x$ over the integers without guessing?

Question

I have no idea how to deal with exponential functions like this. I tried guess and check for $x,y = \pm 1$, finding that $$(x, y) = (1, 1), (1, -1)$$ works. I then tried $x = \pm 2, y = \pm 4$ and $x = \pm 4, y = \pm 2$. I find the solution $$(x, y) = (4, -2)$$ I then wrote a program in python, and these are the only 3 solutions for $-10 \leq x, y \leq 10$

Question 1: How would I solve this system without guessing?

Question 2: Are there any other solutions?

Answer 1

$$x^{x+y}=y^4$$ $$y^{x+y}=x$$

Taking the logarithm both sides,

$$(x+y)\log x =4 \log y $$ $$(x+y)\log y = \log x$$

Substituting [2] into [1],

$$(x+y)(x+y)\log y =4 \log y $$

Cancelling out $\log y$ from both sides,

$$(x + y)^2 = 4 \implies x + y = \pm 2$$

By inspection, since $x,y \in \mathbb{Z}$, the only integer solutions are:

$$x=1, y=1$$ $$x=4, y=-2$$ $$x=-4, y=2$$

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1. $x=-4, y=2$ does not satisfy the original equation.

$$(-4)^{-2} = \frac{1}{16}, 2^{-2} = \frac{1}{4}$$

2. The only other solution $x=1, y=-1$ seems to be a happy coincidence, since:

$$(1)^0 = (-1)^4$$ $$(-1)^0 = 1$$

3. The reverse $x=-1, y=1$ happens to be false.

$$(-1)^0 = (1)^4$$ $$(1)^0 \neq -1$$