Suppose that both $p$ and $q$ are odd. Then $p^2+q^2$ is even and the l.h.s. of the equation is divisible by 4. Squares of integers are congruent to 0 or 1 modulo 4 whence the r.h.s. is congruent to 1 or 2 modulo 4. The contradiction shows that one of $p$ and $q$ equals 2; let w.l.o.g. $p = 2$. Squares of integers are congruent to 0 or 1 modulo 3. Obviously $r > 3$ as the l.h.s. is greater than 10. As $r$ is prime, $r$ is not divisible by 3. Thus $r^2 \equiv 1 \pmod 3$, whence $r^2 + 1 \equiv 2 \pmod 3$. Now $2018 \equiv 2 \pmod 3$ implies $p^2 + q^2 \equiv 1 \pmod 3$ and $p^2 = 4 \equiv 1 \pmod 3$ in turn implies $q^2 \equiv 0 \pmod 3$. Hence $q$ is divisible by 3, i.e., $q = 3$. Therefore the l.h.s. is $2018 \cdot 13$. As $2018 \cdot 13 \equiv 3 \cdot 3 = 9 \equiv 4 \pmod 5$, we must have $r^2 \equiv 3 \pmod 5$. But 3 is not a quadratic residue modulo 5.