If $n=1,2$ there are no solutions. Therefore, taking the equation $\mod 8$, we get $5^k\equiv 1 \mod 8$ or $k$ is even, meaning $5^k$ is a square. The equation $2^n-7=x^2$ is well known to have solutions $(x,n)=(1,3) (3,4) (5,5) (11,7) (181,15)$. Thus the answer is $(n,k)=(5,2)$.

can you explain how to solve $2^n-7=x^2$?

Oh, it is Ramanujan–Nagell equation. I see

i3435 wrote: If $n=1,2$ there are no solutions. Therefore, taking the equation $\mod 8$, we get $5^k\equiv 1 \mod 8$ or $k$ is even, meaning $5^k$ is a square. The equation $2^n-7=x^2$ is well known to have solutions $(x,n)=(1,3) (3,4) (5,5) (11,7) (181,15)$. Thus the answer is $(n,k)=(5,2)$. Hi, what is the motivation of taking mod 8?

@Above $8 = 2^3$ So if $n \geq 3$, it will vanish mod $8$