Showing there are no integer solution to equation $\;2^x = 4y+3$ I am stuck on this problem and I'm not sure how to approach it. Can anyone help me out with figuring how to approach the proof?
My task is to:

Prove that it is impossible to find integers $\,x,\, y\,$ such that $\;2^x = 4y + 3$. 

I assumed a proof by cases would be the way to go?
Any input? Thanks in advance!
 A: Proof-By-Cases - Sketch: 
We consider $x \in \mathbb{Z}$. For all $x \in \mathbb{Z}$:


*

*$x > 0$

*$x = 0,\;$ or

*$x < 0$


$(1)$ For non-negative integer $x (x >0)$: Show the left hand side will always be even, except when $x = 0$, and the right hand side will always be odd, regardless of the integer value of $y$.
(I.e. all positive integral powers of $2$ are even, but $4y+3 = 2\cdot 2 y + 2 + 1 = 2(2y+1) + 1$ must be odd, regardless of the value of $y$.)
$(2)$ Then consider the case $x = 0$: $\;2^0 = 1 \neq 4y+3 = 2(2y+1) + 1$, whatever the integer value of $y$.
$(3)$ For negative integers $x (x < 0):$ the left-hand side will not be an integer $\left(\text{e.g.,}\;\; 2^{-2} = \dfrac 14\right),\;$ while the right hand side will always be an integer, regardless of the value of integer $y$. Hence the equation has not solution in integers in this case, either.

And hence we conclude there are no integer solutions for $x, y$ satisfying the equation: $$2^x = 4y + 3$$
A: $2^x$ is even and $4 y+3$ is odd...
