Note that it suffices to show \[ n\triangleq \left\lceil \frac{p^q+q^p+1}{pq}\right\rceil \]is odd. To that end, let $p^q+q^p+1 = k\cdot pq+r$ where $0\le r\le pq-1$. Then it is clear that $n=k+1$, so it suffices to show $k$ is even. Now, check that as $p\ne q$ are odd primes, $p^q+q^p+1\equiv q+1\pmod{p}$ and $p^q+q^p+1\equiv p+1\pmod{q}$. Combining, it is not hard to recover (using e.g. CRT) that $p^q+q^p+1\equiv p+q+1\pmod{pq}$. Thus, $r=p+q+1$. Now, $k\cdot pq = p^q+q^p+1.- (p+q+1)\equiv 0\pmod{2}$, forces $k$ to be even, as claimed.

it is very easy notice that the upper and lower are odd then if $p^q+q^p-pq+1$ divisible by $pq$ then it will be odd otherwise it is true $p^q+q^p+1 \equiv 0(mod pq)$ $p^q+q^p+1 \equiv 0(mod p)$ $p^q+q^p+1 \equiv 0(mod q)$ $p+1 \equiv 0(mod q)$ $q+1 \equiv 0(mod p)$ we get contradicition.