For the impatient: You can skip almost all of this post, and go to the last $4$ lines.
The informal approach described by the OP almost works, but needs a little tweaking. We start informally, but end up with a quick formal proof. I want to tell the story in a roundabout way, in order to begin to convey the thinking.
Imagine $x$ large positive. It is clear that $\sqrt{x^2-1}$ is smaller, but not much smaller, than $x$, so $\frac{x^2}{\sqrt{x^2-1}}$ is bigger than $x$, and therefore $\frac{x^2}{\sqrt{x^2-1}}-x >0.$ This is not enough to tell us the limit. We need to answer, at least roughly, the question: $\sqrt{x^2-1}$ is how much smaller than $x$?.
Maybe smaller by $u$, where $u$ is small? How small? We have $\sqrt{x^2-1}=x-u,$ and therefore $x^2-1=x^2-2ux+u^2.$ or more simply $2ux-u^2=1$.
Since $u$ is small, $2ux$ is very much bigger than $u^2$. So $2ux\approx 1$, and therefore $u\approx \frac{1}{2x}$. This is an excellent approximation. (By the way, it is the tangent line approximation.)
So, informally, $\frac{x^2}{\sqrt{x^2-1}}-x\approx \frac{x^2}{x-\frac{1}{2x}}-x=\frac{1}{2}\frac{1}{x-\frac{1}{2x}}.$
Great, looks like we are finished. It is obvious that the right-hand side is tiny when $x$ is huge. We even have good control over how tiny.
Unfortunately, for very formal purposes, there is a little problem. For note that if $u=1/(2x)$ then $(x-u)^2$, though a wonderful approximation to $x^2-1$, is unfortunately a little bit bigger than $x^2-1$.
Let's fix that, crudely, because crude is good enough. Instead of $1/(2x)$, we will use $1/x$.
$\left(x- \frac{1}{x}\right)^2 =x^2-2+\frac{1}{x^2}.$ For $x$ large, $x^2-2+\frac{1}{x^2}$ is clearly less than $x^2-1$. So now we know that $\frac{x^2}{\sqrt{1-x^2}}<\frac{x^2}{x-\frac{1}{x}}$. That upper bound will do the job.
Proof of the limit result: We now write up a solution, carefully erasing all traces of the thinking that went into it.
It is easy to verify that if $x>1$, then $(x-1/x)^2 . It follows that $0 <\frac{x^2}{\sqrt{x^2-1}}-x<\frac{x^2}{x-\frac{1}{x}}-x=\frac{1}{x-\frac{1}{x}}.$ By Squeezing, it follows that $\lim_{x\to\infty}\left(\frac{x^2}{\sqrt{x^2-1}}-x\right)=0.$