(2,1) or (1,2) work

proof?

Clearly $a\neq b$. WLOG assume that $a>b$. Here $a\geq2$. Since $a^b+b^a$ and $a^a-b^b$ are positive integers, $a^b+b^a\geq a^a-b^b$. Here, since $b\leq a-1$, $$a^{a-1}+(a-1)^a\geq a^a-(a-1)^{a-1}$$which is equivalent to $$a^{a-2}\leq (a-1)^{a-2}$$Obviously this is false when $a\geq3$. Thus $a=2$ and $b=1$. By symmetry $a=1$ and $b=2$ is the only solution when $a<b$. Therefore the solutions are $(a,b)=(1,2),(2,1)$.