We claim all solutions are as follows: $(a,b)=(1,p^k-1),(p^k-1)$ for any $k\ge 1$ and $(a,b)=(2^k+1,2^k-1),(2^k-1,2^k+1)$ with $k\ge 1$.
Now assume $a,b>1$, $a+b=p^k$ and $a\ge b$ without loss of generality. As $a\ge b>1$ we have $(a-1)(b-1)>0\implies ab+1>a+b$. As a result, $a+b\mid ab+1$. We get $p^k=a+b\mid ab+1$. Moreover $a+b\mid ab+b^2$, yielding $a+b\mid b^2-1 = (b-1)(b+1)$. If $p>2$, this forces $a+b\mid b-1$ or $a+b\mid b+1$. This is contradictory since $a>1$. Thus we have $p=2$. From here, since $(b-1,b+1)=1$, we find that $2^{k-1} \mid b+1$, that is $a+b\le 2b+2$. Moreover, if $a=b$ we have $2a=2^i$, forcing $a$ to be even, but then $ab+1$ is odd. Thus $a\ge b+2$. Hence $2b+2\le a+b\le 2b+2$, yielding that in fact $a=b+2$. With this, $2(b+1)=p^k$ and $(b+1)^2=p^\ell$ for some $k,\ell$. From here, we get $(a,b)=(2^k+1,2^k-1)$.