Does this prove that no sequential squares have a ratio of 2?

Another argument: Either $k$ or $k+1$ is even, the square of that is divisible by $4$, the other is odd, so the ratio can't be $2$.

How about solving the equation: $(k+1)^2=2k^2$ for $k\ne 0$?

$$(k+1)^2=2k^2$$

$$k^2+2k+1=2k^2$$

$$k^2-2k-1=0$$

$$k^2-2k+1=2$$

$$(k-1)^2-(\sqrt{2})^2=0$$

$$(k-1-\sqrt{2})(k-1+\sqrt{2})=0$$

Then $k=1+\sqrt{2}$ or $k=1-\sqrt{2}$, neither of them are integers, because you can prove that $\sqrt{2}$ is an irrational number (but knowing that it is not an integer is enough here).

For your method, the only minor mistake you made is:

$${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$

Your proof is OK, but there is no need to invoke the irrationality of $\sqrt2$. It's enough to note that for

$$f(k)={(k+1)^2\over k^2}=\left(1+{1\over k}\right)^2$$

(with $k\not=0$), we have $f(k)\lt1$ if $k\lt0$ and

$$f(1)\gt f(2)=9/4\gt2\gt16/9=f(3)\gt f(4)\gt\cdots$$

Just to be different:

$\frac {(k+1)^2}{k^2} = \frac {k^2 + 2k + 1}{k^2} = 1 + \frac {2k + 1}{k^2}$

So $\frac {(k+1)^2}{k^2} = 2\implies \frac {2k + 1}{k^2} = 1 \implies$

$2k + 1 =k^2$ and as $k \ne 0$ (otherwise $2 = 0$ which isn't true)

$2 + \frac 1k = k$. But if $k \ne \pm 1$ then $\frac 1k$ is not an integer.

And if $k = \pm 1$ then $\frac 1k = k$ and $2 = k -\frac 1k = 0$ which is impossible.

....

But seriously...

$\frac {(k+1)^2}{k^2} = (\frac {k+1}{k})^2 = 2$ implies there is a rational number whose square is $2$ which is well-known to be false, if a perfect and irrefutable proof. And probably the absolute easiest.

If you want to invoke the irrationality of $\sqrt 2$, you can prove a much stronger result, that no two perfect squares $m^2$ and $n^2$ can exist so that $\frac{m^2}{n^2} = 2$. It's as simple as taking the square root of both sides and noting the rationality of one side and the irrationality of the other.

The reason I am bringing this up is because you mentioned in a comment that this was a problem you came up with, so I thought it might be nice to lead you to an even stronger result than the one you conjectured.