Is "Infinity" a valid solution? 
Is the above solution correct?
The point I wanted to confirm is that the expression asymptotically approaches 1 as x approaches infinity.
(In other words, as the limit of the fraction as x approaches infinity  is 1).
So, for k=1, does the fraction have one solution or two?
 A: There are two issues here. One is what is the range of allowable values that $x$ can have to qualify as potential "solutions" or not. The other is what it means for a value to be the solution of an equation.
For the first issue, that range is almost surely $\mathbb{R}$, the real numbers, and since $\infty$ is not a real number, it does not belong as a solution. That's why, if you note, the intervals involving $\infty$ are always given as having "open" symbols on the side where $\infty$ is put - a closed interval that explicitly includes $\infty$, e.g. $[-\infty, -7 - 4\sqrt{3}]$, would not be an interval of real numbers alone.
That then makes one wonder, to what does $\infty$ belong? The answer is it is an extended real number, which is a different number system than $\mathbb{R}$. It may be denoted $\overline{\mathbb{R}}$, though I prefer $\mathbb{R}_\mathrm{ext}$. The extended line is not the real line, and moreover requires great care when it comes to interpreting expressions thereupon: in particular, expressions like $\infty - \infty$ and $\frac{\infty}{\infty}$ are undefined, even while we do have $\frac{1}{\infty} = 0$, and even $\frac{1}{0}$ is still undefined.
So now for what constitutes a solution. Would you say that $x = 2$ is a solution of
$$\frac{x - 2}{(x - 2)(3 - x)} = 1$$
? Actually, directly as stated, the answer is no: the expression on the left is undefined when $x = 2$. It is only after cancellation that you introduce this as a solution to the equation.
Likewise, I imagine that you are thinking that in the case of
$$\frac{x^2 + 4x + 3}{x^2 - 4x - 3} = 1$$
we can limit-extend to $x = \infty$. Yes, we can, but that's kinda the same as the previous case: we've essentially changed the equation so that a solution has become introduced that was not there for the original. In the original, the value of the left-hand expression is undefined, even in $\mathbb{R}_\mathrm{ext}$, because the denominator acquires an $\infty - \infty$.
This points to another difference between $\mathbb{R}$ and $\mathbb{R}_\mathrm{ext}$: in the former, $0x = 0$, but in the latter, $0x \ne 0$ in general because when $x = \infty$, we get $0 \cdot \infty$, which is undefined!
