If $p|ab\implies p|a, p|b.$ Since $ab\ge 2.$ Let $q$ be a prime dividing $ab.$ So $q|a,q|b.$ And $q|a^2+b^2.$ So note that $a,b$ share the same set of prime divisors. Hence $$v_q({ab}^{2015})=v_q((a^2+b^2)^m).$$So $$2015\cdot v_q(ab)=m\cdot v_q(a^2+b^2).$$ Clearly, since we have the same set of primes, we get $m|2015.$ And $m\ne 2015$ because by AM-GM $$a^2+b^2\ge 2ab >ab.$$So $$m< 2015.$$ Let $$a=q^k\alpha, (p,\alpha)=1,~~b=q^y\beta, (p,\beta)=1.$$Then $$v_q((a^2 + b^2)^m)=m\cdot \min(y,k).$$And $$v_q((ab)^{2015} )= 2015 \cdot (y+k).$$ So we get $$m\ge 2015.$$ So there is no $m.$ Note: If $a,b = 1$ then it's np since both would $a,b=1$ which is not possible.