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

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