For all solutions of n and m that satisfy.

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3\times2^{m}+1=n^{2}

For some positive integers m and n.
 
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3*2^3+1=5^2
 
A nice way to solve this is to factor:
3\times 2^m = (n-1)(n+1)
Now clearly one factor on the right is a power of 2 since the factor 3 can only occur in one of them. Split it into cases depending on which it is and you should get the answers.
 
of (n-1)(n+1) one factor has to be a power of 2 and the other 2*3=6. so the two possibiltites are 4*6 and 6*8.

So all answers are 3*2^3+1=5^2 and 3*2^4+1=7^2.
 

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