This solution unfortunately came out quite a bit longer than I expected (certainly not as easy as checking Sage), but it is entirely elementary and self contained. I'm sure there is a significantly shorter proof, but the idea is just to bash the equation using the $pqrs$ lemma and the parametrization for primitive Pythagorean triples.
As you have shown, $p$ must be even. Therefore by the parametrization of Pythagorean triples, there exist natural numbers $a,b,c,d$ such that
$q = a^2 - b^2, p = 2ab, n=a^2+b^2$
$q-p = c^2 - d^2, p = 2cd, m =c^2+d^2$
Equivalently, we are given four integers $a,b,c,d$ such that $\gcd(a,b)=\gcd(c,d)=1, ab=cd, a^2-b^2=c^2-d^2+2cd$
The $pqrs$ lemma states that given four positive integers $a,b,c,d$ such that $\gcd(a,b)=\gcd(c,d)=1, ab=cd$, there exist 4 pairwise coprime positive integers $p,q,r,s$ such that $a=pq, b=rs, c=pr, d=qs$. (The proof is simple: define $p=\gcd(a,c), q=\gcd(a,d)...$).
Using it here (this is a slight abuse of notation, as these $p,q,r,s$ are not related to the original $p,q$ but whatever), we get the equation
$p^2q^2 - r^2s^2 = p^2r^2 - q^2s^2 +2pqrs$
$p^2(q^2-r^2) - 2pqrs +s^2(q^2-r^2)=0$
Excluding the case $q=r=1$ (which you can easily check does not lead to a solution), this is a quadratic equation in $p$. For a quadratic equation to have a solution in integers, the discriminant must be a square, therefore $s^2(q^2r^2-(q^2-r^2)^2)$ is a square. Equivalently (as obviously $s \neq 0$),
$(r^2 + rq - q^2)(q^2 + rq - r^2)$ is a square. Notice that this is a product of two coprime integers (their $\gcd$ must divide their sum, which is $2rq$, but they are each coprime to $2, r, q$), so since their product is a square, each of them individually must be a square.
$r^2 + rq - q^2 = e^2$
$q^2+ qr - r^2 = f^2$
We will assume without loss of generality that $r > q$.
Adding and subtracting we get the equivalent equations
$2qr = e^2 + f^2$
$2(r^2 - q^2) = e^2 - f^2$
Notice from the first equation that $e,f$ must both be odd, and then looking $\mod 4$ we find that $q,r$ must both be odd as well. Using this, we define the four positive integers $x = \frac{e+f}{2}, y= \frac{e-f}{2}, z = \frac{q+r}{2}, t = \frac{q-r}{2}$, and we can rewrite our equation in these terms:
$z^2 - t^2 = x^2 + y^2$
$2tz = xy$
One of $x,y$ must be even, say $x = 2x'$. Then $x'y = tz$, and again using the $pqrs$ lemma we get four positive pairwise coprime integers $p', q', r', s'$ such that $x' = p'q', y = r's', z = p'r', t = q's'$, and the first equation can be written as
$p'^2 r'^2 - q'^2 s'^2 = 4p'^2 q'^2 + r'^2 s'^2$
$p'^2 (r'^2 - 4q'^2) = s'^2(q'^2 + r'^2)$
Therefore $(q'^2 + r'^2)(r'^2 - 4q'^2)$ is a nonzero square. We will show using the method of infinite descent that this cannot happen. Notice that $\gcd(r'^2 - 4q'^2, r'^2 + q'^2) | 5q'^2$, but they are both clearly coprime to $q'$ so $\gcd(r'^2 - 4q'^2, r'^2 + q'^2) | 5$.
Case 1: the gcd is equal to 1. In this case,
$p'^2 = q'^2 + r'^2, r'^2 = s'^2 + (2q'^2)^2$
A computation modulo 8 shows that r' must be odd, so these are two primitive Pythagorean triples. Using the parametrization we get numbers $x',y',z',t'$ such that $r' = z'^2 - t'^2, q' = 2t'z', p' = t'^2 + z'^2$ and $r' = x'^2 + y'^2, 2q' = 2x'y', s' = x'^2 - y'^2$
In other words we have found numbers $x',y',z',t'$ such that
$2t'z' = x'y'$
$z'^2 - t'^2 = x'^2 + y'^2$
Which is exactly what the original $x,y,z,t$ satisfied! So we can simply replicate the process we used to generate a smaller pair $q', r'$ such that $(q'^2 + r'^2)(r'^2 - 4q'^2)$ is a nonzero square. This means that what must eventually occur is
Case 2: the gcd is equal to 5.
Therefore there exist $x,y$ such that $q'^2 + r'^2 = 5x^2, (r'-2q')(r'+2q')=5y^2$.
Denoting $k = \gcd(r'-2q', r'+2q')$, we have two cases.
Case 2a) $r'-2q' = 5k\cdot \alpha^2, r' + 2q' = k \cdot \beta^2$ where $y=k\alpha \beta$, and $\alpha, \beta$ are coprime.
Then $r' = k\cdot \frac{5 \alpha^2 + \beta^2}{2}, q' = k\cdot \frac{\beta^2 - 5 \alpha^2}{4}$. Substituting this information into our equation we get that
$r'^2 + q'^2 = (\frac{k}{4})^2 \cdot (125 \alpha^4 + 30 \alpha^2 \beta^2 + 5 \beta^4) = 5x^2$
$\beta^4 + 6\alpha^2 \beta^2 + 25\alpha^4 = (\frac{4x}{k})^2$
$(\beta^2 + 3\alpha^2)^2 + (4\alpha^2)^2 = (\frac{4x}{k})^2$
Now, if both $\alpha, \beta$ are odd then we get $(\frac{\beta^2 + 3\alpha^2}{4})^2 + (\alpha^2)^2 = (\frac{x}{k})^2$, which is a Pythagorean triple with 2 odd numbers, which is impossible. Therefore, exactly one of $\alpha, \beta$ is even so $\beta^2 + 3\alpha^2$ is odd and $(\beta^2 + 3\alpha^2)^2 + (4\alpha^2)^2 = (\frac{4x}{k})^2$ is a primitive Pythagorean triple. One final use of the parametrization give us two positive integers $m,n$ such that
$4\alpha^2 = 2mn, \beta^2 + 3\alpha^2 = m^2 - n^2$
$2\alpha^2 = mn$, so one of $m,n$ is a square and the other twice a square.
Case 2a)i. $m = 2u^2, n = v^2, \alpha = uv$. Substituting, we get
$\beta^2 = 4u^4 - 3u^2v^2 - v^4 = (4u^2 - v^2)(u^2 + v^2)$, so we managed to create an even smaller pair $u,v$ which fits equation what we wanted from $q', r'$. As we cannot infinitely descend in this way we must eventually arrive at
Case 2b)ii. $m = u^2, n = 2v^2, \alpha = uv$. Substituting, we get
$\beta^2 = u^4 - 3u^2v^2 - 4v^4 = (u^2-v^2)(u^2 + 4v^2) = (u-v)(u+v)(u^2 + 4v^2)$
$\gcd(u^2 - v^2, u^2 + 4v^2)|5$. If the $\gcd$ is one they are both squares and we get $v = gh, u = g^2 - h^2$ from the second square and $(g^2 - gh - h^2)(g^2 + gh - h^2)$ is a square. We can use the same techniques to eventually arrive at a smaller pair $q', r'$.
If the $\gcd$ is five then we have another case which is similar (unfortunately I do not have time to write this up).
EDIT: I was a bit lazy here, it turns out that the $\gcd$ cannot be $5$: looking $\mod 4$ using the fact that $u^2 - v^2$ is 5 times a square we find that $v$ is even and $u$ is odd, but then $u^2 + 4v^2 \equiv 1 (\mod 8)$ which is not 5 times a square.
Case 2b) where $r'-2q' = k\cdot \alpha^2, r' + 2q' = 5k \cdot \beta^2$ where $y=k\alpha \beta$, and $\alpha, \beta$ are coprime is completely identical to Case 2a).
SUMMARY
We showed that to prove the problem it is sufficient to show that there do not exist a pair of coprime positive integers $q', r'$ such that $(q'^2 + r'^2)(q'^2 - 4r'^2)$ is a nonzero square. Then, we showed that given any such pair we can find a smaller pair of coprime positive integers such that the relevant product is a square. However, we cannot descend forever, which means that no such pair $q', r'$ exists, which proves the question.
Best Answer
You actually want it the other way around: if $a^2+b^2=c^2$ then $a+b+c|abc$. That you can prove very quickly from the general form of primitive Pythagorean triples $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.