The following passage can be found in Burton's Elementary Number Theory:
"Assume that $a$ is odd. Because the square of any odd integer is congruent to 1 modulo 8, we see that for the congruence $x^2 \equiv a \pmod {2^n},$ where $n\geq 3,$ to be solvable $a$ must be of the form $8k + 1.$"
The only thing I know here is that the square of any odd integer is congruent to 1 modulo 8. Using this, how to show the last part?
Consider any solution, where $x^2 = a + j 2^n$ for some integer $j$; $a$ is odd, so $x^2$ is odd, so $x$ is odd. Thus $a + j 2^n$ is the square of an odd integer, and is congruent to $1$ mod $8$. But $2^n$ is also divisible by $8$ since $n \ge 3$, so $a$ is congruent to $1$ mod $8$.