Is my proof correct? Proof: $n^{2}$ is odd then $n$ is odd.

Very well done. Yes, indeed, proving $n\text{ is even}\implies n^2\text{ is even}$ is much easier than the original statement.

Alternatively, if $n^2$ is odd, then $n^2-1$ is even. It can be expressed as: $$n^2-1=(n-1)(n+1).$$ If $n$ is even, $n-1$ and $n+1$ are both odd, whose product is also odd and it contradicts the given condition ($n^2-1$ is even). Hence, $n$ is odd.

Your idea is correct, but it might be better to write the proof this way:

We prove the contrapositive statement, that if $n$ is even then $n^2$ is even. Since $n$ is even, $n=2k$ for some integer $k$. Then $n^2=4k^2=2(2k^2)$ with $2k^2$ an integer, so $n^2$ is even.

Notice that in your proof, you did not mention that $m$ or $2m^2$ are integers, which are important details. The last two biconditional arrows also obscure the picture, especially when the nature of $m$ is not specified. (You only need one direction in the proof, and the other direction might not even hold in some cases.)