If $(x,y) = (2,6)$ then $f(x,y) = 20$, and as $y$ increases above $6$, $f(x,y)$ increases.
To prove this rigorously, let us prove by induction that $$2^{X-1}\gt X+1\tag1$$ for $X\ge 6$.
The base case : $2^{6-1}=32\gt 7=6+1$.
Supposing that $(1)$ holds for some $X$ gives
$$2^{(X+1)-1}=2\cdot 2^{X-1}\gt 2(X+1)=X+X+2\gt (X+1)+1\quad\blacksquare$$
Using $(1)$, we get, for $y\ge 6$,
$$f(2,y+1)-f(2,y)=2(2^{y-1}-y-1)\gt 0$$
If $x \geq 3$ and $y \leq x$ then $f(x,y) < 0$.
$f(x,1)=-2\lt 0$.
To prove that $f(x,2)\lt 0$ for $x\ge 3$, let us prove by induction that $$2^X\gt X^2-X-2\tag2$$ for $X\ge 4$. The base case : $2^4=16\gt 10=4^2-4-2$. Supposing that $(2)$ holds for some $X$ gives $$\begin{align}2^{X+1}&=2\cdot 2^X\\&\gt 2(X^2-X-2)\\&=(X+1)^2-(X+1)-2+X(X-3)-2\\&\ge (X+1)^2-(X+1)-2+4(4-3)-2\\&\gt (X+1)^2-(X+1)-2\quad\blacksquare\end{align}$$ Also, it is easy to see that $(2)$ holds for $X=3$. Using this, we get, for $x\ge 3$, $$f(x,2)=(x^2-x-2)-2^x\lt 0$$
To prove that $f(x,y)\lt 0$ for $3\le y\le x$, we use the fact that $y=\frac{\ln x}{x}$ is decreasing for $x\gt e$. We see that $\frac{\ln y}{y}\ge \frac{\ln x}{x}$ for $3\le y\le x$ from which $x^y-y^x\le 0$ follows. It follows from this that we have, for $3\le y\le x$, $$f(x,y)=x^y-y^x-x-y\le 0-x-y\lt 0$$
If $x \geq 3$ and $y=x+1$, then $f(x,y)>0$, and as $y$ increases above $x+1$, $f(x,y)$ increases.
This answer proves that $f(x,y)\gt 0$ for $x\ge 3$ and $y\ge x+2$.
Let $y=x+a$ where $a\ge 2$ is an integer. Then,
$$\frac{f(x,x+a)}{x^x}=x^{a}-\left(1+\frac ax\right)^x-\frac{2}{x^{x-1}}-\frac{a}{x^x}\gt x^a-e^a-\frac 29-\frac{a}{x^x}:=g(x)$$
Since $g'(x)=ax^{a-1}+\frac{a+a\ln x}{x^x}\gt 0$, $g(x)$ is increasing with $g(3)=3^a-e^a-\frac 29-\frac{a}{27}:=h(a)$.
We have $h'(a)=3^a\ln 3-e^a-\frac{1}{27}$ and $h''(a)=3^a(\ln 3)^2-e^a\gt 0$ with $h'(2)=9\ln 3-e^2-\frac{1}{27}\gt 0$ and $h(2)=9-e^2-\frac 29-\frac{2}{27}\gt 0$.
So, $h(a)\gt 0$ for $a\ge 2$. It follows from this that $g(x)\gt 0$ for $x\ge 3$.
Therefore, $f(x,x+a)\gt 0$ follows.
