Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. Today my friend asked a question for help:

Find all solutions of $2^x=5^y+3$ for non-negative integers $x,y$.

It is obvious that the solutions are $(x,y)=(2,0),(3,1),(7,3)$, and I think there is no more solution. However, we can't prove that these are the only solutions. We have already tried to mod many numbers and still get "may" possible solutions other than the $3$ solutions I have written. I have surfed the Net and still can't find solutions. I hope you guys can help my friend solve. Thank you very much!
 A: In case anyone wants to learn what is going on in the solution, here are my earlier examples and the person who discovered the method (answer at the first link):
http://math.stackexchange.com/questions/1551324/exponential-diophantine-equation-7y-2-3x
http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7
http://math.stackexchange.com/questions/1941354/elementary-solution-of-exponential-diophantine-equation-2x-3y-7/1942409#1942409 
http://math.stackexchange.com/questions/1946621/finding-solutions-to-the-diophantine-equation-7a-3b100/1946810#1946810 
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
The diophantine equation $5\times 2^{x-4}=3^y-1$
Equation in integers $7^x-3^y=4$ 
Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. 
A: From WolframAlpha here we get
$$x = \frac{\log{5^y + 3}}{\log{2}}\quad\text{for}\quad
 5^y\in\mathbb{Z}\quad\land\quad5^y >=-2$$
The restrictions are overkill but simply indicate $y\in\mathbb{Z}$
Applying this in a spreadsheet for $-15000\le y \le 15000$ we get only $3$ pairs.
$$(2,0)\quad
(3,1)\quad
(7,3)\quad$$
