The goal: Prove that there is no integer $k$ such that ${(k+1)^2\over{k^2}}=2$.
My proof: If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$, and if $k$ is an integer, $k+1$ is also an integer. This implies that $\sqrt2$ is a rational number, which is also provably false.
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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.
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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 $2$.
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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
$$f(1)\gt f(2)=9/4\gt2\gt16/9=f(3)\gt f(4)\gt\cdots$$
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And yet another way... if $k\ge3$ then $$2\le\frac23k\ ,\qquad 1\le\frac19k^2$$ so $$(k+1)^2=k^2+2k+1\le k^2+\frac23k^2+\frac19k^2=\frac{16}9k^2<2k^2\ .$$ This leaves $k=1,2$ as the only options, and it's easy to check that they don't work either.
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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.
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While the other answers are focused on how this can be proven easier I am focusing on the question if your proof is correct.
Your proof contains an error:
If ${(k+1)^2\over{k^2}}=2$, then ${{k+1}\over{k}}=\sqrt2$
However ${{k+1}\over{k}}$ may also be $(-\sqrt2)$ in this case.
If you however argue that neither $\sqrt2$ nor $(-\sqrt2)$ are rational numbers your proof will work.
On
No sequential integer squares will ever have a ratio of two. In fact, no integer squares will ever have a ratio of two. That is to say, no integer square will ever be exactly twice the size of another.
Start by assuming $a^2=2b^2$. The quantity $b^2$ has the prime factorization of $list*list$, where "list" is the prime factorization of $b$, multiplied in long form. If $b$ were 10, for example, the element "list" would be replaced with $5*2$.
Because there are two of each prime factor from $b$, we deduce that $b^2$ has an even number of prime factors. The same argument may show that $a^2$ will also have an even number of prime factors.
Obviously, $a^2$ cannot have the prime factorization $primelist*primelist*2$, as this is an odd number of prime factors. Contradiction.
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:
$${(k+1)^2\over{k^2}}=2\Leftrightarrow {{k+1}\over{k}}=\pm \sqrt2.$$