http://en.wikipedia.org/wiki/Ramanujan-Nagell_equation

This is much easier than the Ramanujan-Nagell equation. Exponentials and polynomials share the property of being periodic modulo a prime.

n = { 3,4,5,7,15 }

Is it just me, or is the Ramanujan-Nagell equation and this problem very different problems.

I don't know, this site tell me: http://mathworld.wolfram.com/RamanujansSquareEquation.html

That is not the equation we're looking at.

$ 2^n$ modulo 7 has period of 3. $ n^2$ modulo 7 has period of 7. So $ 2^n-n^2$ modulo 7 has period of 21. It is easy to check $ n=1$ to 21.

Note that $2^n$ is $1,2,4 \pmod 7,$ and $n^2 \equiv 0,1,2,4 \pmod 7,$ now note that we need both to be either $1,2,4 \pmod 7.$ Now we can take casework. If both are $1\pmod 7$ then $n \equiv 0 \pmod 3,$ and $1,6 \pmod 7,$ so $n$ can be either $15 \pmod {21},$ and $6 \pmod {21}.$ Now for the case of $2\pmod 7,$ we know that $n \equiv 1 \pmod 3,$ and $3,4 \pmod 7,$ thus we can conclude that $n \equiv 4, 10 \pmod {21}.$ Now for the last case if they are both $4\pmod 7,$ then we have that $n \equiv 2,5 \pmod 7$ and $n \equiv 2\pmod 3,$ we can see that the solutions are $2, 5 \pmod {21}$ thus all natural numbers $n$ are in the form of $\boxed{2,4,5,6,10,15 \pmod {21}}.$