We shall show more generally that $am^2 + bm + c \equiv 0 \pmod{2^n}$ has a solution for all $n$ whenever $b$ is odd and $a$ or $c$ is even. For $n = 1$, take $m = 0$ if $c$ is even and $m = 1$ if $c$ is odd. Now suppose the claim is true for $n$. If $c$ is even, then by assumption, the congruence $2at^2 + bt + c/2 \equiv 0 \pmod{2^n}$ has a solution in $t$. Letting $m = 2t$, we get $am^2 + bm + c = 2(2at^2 + bt + c/2) \equiv 0 \pmod{2^{n+1}}$. If $c$ is odd, then $a$ is even and thus $a+b+c$ is even; hence by assumption, the congruence $2at^2 + (2a+b)t + (a+b+c)/2 \equiv 0 \pmod{2^n}$ has a solution in $t$. Letting $m = 2t+1$, we get $am^2 + bm + c = 2(2at^2 + (2a+b)t + (a+b+c)/2) \equiv 0 \pmod{2^{n+1}}$. This completes the proof of the claim by induction.