Let p be a fixed odd prime. We prove the following results for positive integer solutions $(x,m,n)$ of the equation $(*)\ x^2=2^m+p^n$. (i) If $p \equiv 3 \pmod 8$, then $(*)$ has only the solution $(p,x,m,n)=(3,5,4,2)$. (ii) If $p \equiv 5 \pmod 8$, then $(*)$ has only the solution $(p,x,m,n)=(5,3,2,1)$. (iii) If $p \equiv 7 \pmod 8$, then $(*)$ has at most one solution $(x,m,n)$, except for $p=7$, $(x,m,n)=(3,1,1)$ and $(9,5,2)$. Moreover, if $p=2^q-1$ is a Mersenne prime with $p>7$, where q is an odd prime with $q>3$, then $(*)$ has exactly one solution $(x,m,n)=(2^q+1,q+2,2)$. If $p \equiv 7\pmod 8$, p is not a Mersenne prime and either $p<1.5\times 10^{12}$ or $p>C$, where C is an effectively computable absolute constant, then $(*)$ has only the solutions $p=a^2-2$, where a is an odd positive integer, $(x,m,n)=(a,1,1)$. (iv) If $p \equiv 1\pmod 8$ with $p \ne 17$, then $(*)$ has at most two solutions $(x,m,n)$.