I'm interested in the congruence problem $$(1 + 2^m)^{2x} \equiv 1+ 2^{m+1} \pmod{2^{2m+2}}\,,\quad m \geq 2\,.\quad(\dagger)$$
Though I can solve it for various small values of $m$, I still haven't found a proof that a solution exists for every such $m$.
For context, I'm reading a paper and thinking about the possibilities of extending its results. The author himself claims that extending the theorem in a certain direction is trivial, and part of this is because he claims that $(\dagger)$ always has a solution for each $m \geq 2$.
I have tried so far to expand $(1 + 2^m)^{2x}$ using the binomial theorem and reducing the problem to the quadratic equation $$2^mx^2 + (1-2^{m-1})x-1\equiv 0 \pmod{2^{m+1}}\,,\quad m\geq 2\,.\quad (\ddagger)$$ but that didn't helped much either.
I'm looking for hints to prove that $(\dagger)$ or equivalently $(\ddagger)$ aways has a solution $x$ for every such value of $m$.
If we choose $x$ such that $$ (1-2^{m-1})x\equiv 1+2^m\;(\text{mod}\;2^{m+1}) $$ then $2^mx^2$ and $(1-2^{m-1})x-1$ are both odd multiples of $2^m$, hence $$ 2^mx^2 + \bigl((1-2^{m-1})x-1\bigr) $$ is a multiple of $2^{m+1}$.