For all solutions of n and m that satisfy.

[icystrike]

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

For some positive integers m and n.

 

[robert2734]

3*2^3+1=5^2

 

[rasmhop]

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.

 

[robert2734]

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.