We consider the case $p\geq 3$. For $p=2$ only 1 part is different and we make a remake when applicable. Let $d=\gcd(m,n)$ and $m=ud,n=vd$. Without loss of generality we also assume that $m\leq n$.
Theorem
We will show the following: If $u,v \not \equiv 0 \pmod 2$, then
$$
\gcd(p^m+1,p^n+1) = p^d+1
$$
otherwise exactly one of $u,v$ is even and
$$
\gcd(p^m+1,p^n+1) = 2
$$
(If $p=2$, then the latter $\gcd$ is $1$ instead.)
Euclidean algorithm
The proof follows a series of $\gcd$ reductions as follows:
(a) If $m=n$ or $m=0$: then $\gcd(p^m+1,p^n+1)= p^m+1$ and $2$ respectively. (Latter is $1$ if $p=2$.)
(b) If $2m \leq n$: Let $r=n-2m$, then
$$
\begin{align*}
\gcd(p^m+1,p^n+1) &= \gcd(p^m+1,(p^n+1) - p^{n-m}(p^m + 1) + p^{n-2m}(p^m+1)) \\
&=\gcd(p^m+1,p^{n-2m}+1) \\
&= \gcd(p^m+1,p^r+1)\\
&= \gcd(p^r+1,p^m+1)
\end{align*}
$$
Update $(m,n)=(r,m)$.
(c) Finally, since $m\leq n$, we have $m < n < 2m$. Let $r=2m-n > 0$, then
$$
\begin{align*}
\gcd(p^m+1,p^n+1) &= \gcd(p^m+1,p^{n+r}+p^r) \\
&= \gcd(p^m+1, p^{2m}+p^r) \\
&= \gcd(p^m+1, (p^{2m}+p^r) - p^m(p^m+1) + (p^m+1)) \\
&= \gcd(p^m+1,p^r+1)\\
&= \gcd(p^r+1,p^m+1)
\end{align*}
$$
Update $(m,n)=(r,m)$.
Note: We may check that $\gcd(u,v)=1$ is maintained for each iteration.
Observe that for (c), $r = 2m-n < 2m-m = m < n$ so both (b) and (c) strictly reduces $(m,n)$ to smaller values. Hence if we always have $m\neq n$ then $m=0$ at some point. This shows that the sequence terminates.
Proof
We start with
$$
\gcd(p^m+1,p^n+1) = \gcd(p^{ud}+1,p^{vd}+1)
$$
and apply the reductions iteratively. At each iteration that is not (a), since $r=n-2m$ or $r=2m-n$, we have
$$
\gcd(m,r) = \gcd(m,\pm(n-2m)) = \gcd(m,n) = d
$$
Hence after update of $(m,n)$ we still have the form
$$
\gcd(p^{ud}+1,p^{vd}+1)
$$
i.e. $\gcd(m,n)=d$ is maintained.
Moreover, the parity of $u,v$ is swapped but otherwise unchanged. We see this immediately for the new $v$ since it equals the old $u$. For the new $u=r/d$, we have
$$
\begin{align*}
r &= \pm(n-2m) =\pm(vd-2ud)\\
r/d &= \pm(v-2u) \equiv v \pmod 2
\end{align*}
$$
So parity of new $u$ equals parity of old $v$.
Now we go back to the original case of $u,v \not\equiv 0\pmod 2$. Since both $u,v$ has odd parity, $u\neq 0$ and we cannot terminate at the case where $m=0$. This means we terminate at the case where $m=n$. Each update iteration maintains $\gcd(u,v)=1$ so this is only possible if $m=n=d$. Hence the result is
$$\gcd(p^m+1,p^n+1)=\gcd(p^d+1,p^d+1)=p^d+1$$
On the other hand, $u\not \equiv v \pmod 2$, so we cannot have the case $m=n$. Hence we must terminate at $m=0$, which gives us
$$\gcd(p^m+1,p^n+1)=\gcd(p^0+1,p^n+1)=2$$
(If $p=2$ then $p^n+1$ is odd so the $\gcd$ is 1.)
