Incorrect method to find a tilted asymptote

By definition, the function $f(x)$ has as an oblique (slanted) asymptote the straight line $y=\alpha x+\beta$ if: $$ \lim_{x\to+\infty}\big(f(x)-(\alpha x+\beta)\big)=0 $$ or if: $$ \lim_{x\to-\infty}\big(f(x)-(\alpha x+\beta)\big)=0 $$ It easy to check that while your result $y=x-1$ satisfies the above definition, the line $y=x+1$ does not, since: $$ \lim_{x\to\pm\infty}\big(f(x)-(x+1)\big)=-2 $$ P.S.: What you have actually shown via your second argument is that: $$ \lim_{x\to\pm\infty}\frac{f(x)}{x+1}=1 $$ which is however irrelevant to the notion of oblique (slanted) asymptote.

The trouble is that the numerator in $$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$ stays large, so a small error in the denominator is magnified. Better to write $$ \frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots $$ and multiply; the infinite series indicated converges for $|x| > 2.$

$$ \left( x+1-\frac{6}{x} \right) \left( 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots \right) = x - 1 - \frac{4}{x} + \frac{8}{x^2} + \cdots $$

$\frac{y}{x+1} \;\to\; 1$ indeed, but:

$$y - (x+1) = \frac{x^2 + x - 6 - x^2 - 3 x - 3}{x+2} = \frac{-2 x - 9}{x+2} \quad \to \quad -2$$

which is why the slope of $x+1$ is correct, but its intercept is off by $-2$.