For $ n\ge3$ we have $ x^2 + y^2 + z^2 + t^2 = 2^n$, and since $ x^2\equiv0,1,4\pmod8$, we know that $ 2|\gcd(x,y,z,t)$. So cancelling a factor $ 4$ from both sides we get at least one of $ x,y,z,y$ odd, and by the above this implies we need $ 0 + 0 + 0 + 1 = 1$ or $ 1 + 1 + 1 + 1 = 4$ modulo 8, but the only way to have $ 2^n\equiv1,4\pmod8$ is $ n = 1,2$ respectively. The first has no solutions, the latter only $ 1^2 + 1^2 + 1^2 + 1^2 = 2^2$, so the final answer is: there are no solutions for $ n$ odd, and for $ n = 2k$ there is a unique solution $ \left(2^{k - 1}\right) + \left(2^{k - 1}\right) + \left(2^{k - 1}\right) + \left(2^{k - 1}\right) = 2^{2k}$. [mod team: solution rewritten readably]