5^p * 43^q = (r-1) * (r^2+r+1).

r^2+r+1 = (r-1)*(r+2)-3, which then means GCD(r^2+r+1, r-1) = GCD(r-1, 3). But then that means any common divisor is a divisor of 3.  However the prime factors of (r-1) * (r^2+r+1) are limited to 5 and 43.  Therefore r^2+r+1 and r-1 are coprime.  From this 5^p and 43^q are in some order r-1 and r^2+r+1 since we are working over the integers.

Now calculating r^2+r+1 mod 5 will have values of 1, 2, or 3.  None of these are zero.  And since 5^p * 43^q = (r-1) * (r^2+r+1) is factored over the integers we must conclude that r-1 = 5^p and r^2+r+1 = 43^q.

Then r = 5^p+1.  Substitute this to create 25^p + 3*5^p + 3 = 43^q.  Taking both sides mod 100: The left side equals 43 exactly when p=1 or the left side equals 3 for all p>=2.  The right side cycles through 43, 49, 7, 1 for q=1, 2, 3, 0 (mod 4).  These match only when p=1 and q=1 mod 4.

So the single potential solution comes from p=1.  Then 25^1 + 3*5^1 + 3 = 43 = 43^1, which makes q=1.  Then r = 5^1 + 1 = 6.  Final check: 5^1 * 43^1 = 6^3 - 1 resolves to 215 = 215, which checks out.  Then the only solution is (p,q,r) = (1,1,6).