I am very new to proofs, and I would like to make sure that I am doing them correctly. I am attempting to learn them outside of class, so any comments about proof style or general logical thinking processes would be greatly appreciated! Sorry for the poor formatting as well, I was having some difficulties transferring this proof in a more readable manner.
Proof
For every integer x, if x is odd, then there exists an integer y such that x^2 = 8y+1.
Let x be an odd integer and y be arbitrary.
x = 2y+1
This implies that
x^2 = (2y+1)^2
x^2 = 4y^2+4y+1.
If x^2 = 8y+1 then,
8y+1 = 4y^2+4y+1.
0 = 4y^2-4y
0 = 4y(y-1)
y = 1, y = 0
This implies that there exists an integer y in the set of all integers such that x^2=8y+1.
Q.E.D
You cannot assume what you are trying to prove, so the step $8y+1 = 4y^2+4y+1$ is invalid. Furthermore, you are confusing the $y$ that you defined and the $y$ in the question, which are two different things.
Instead, think about what happens if $m$ (the '$y$' you defined) is odd or even since all integers must be either odd or even. This means you can either have $m=2n$ or $m=2n+1$. After substituting this into $x^2 = 4m^2+4m+1$, try to group the terms so that you have your result in the form $8(\text{something}) + 1$.