Balkan Mathematical Olympiad

For $p=2$ it is directly checked that there are no solutions. Assume that $p>2$. Observe that $N=11^p+17^p \equiv 4 \pmod 8$, so $8 \nmid 3p^{q-1}+1 > 4$. Consider an odd prime divisor $r$ of $3p^{q-1}+1$. Obviously, $r \notin \{3,11,17\}$. There exist $b$ such that $17b \equiv 1 \pmod r$. Then $r|b^pN \equiv a^p+1 \pmod r$,

where $a=11b$. Thus $r|a^{2p}-1$, but $r \nmid a^p-1$, which means that $\operatorname{ord}_r(a)|2p$ and $\operatorname{ord}_r(a) \nmid p$, i.e. $\operatorname{ord}_r(a) \in \{2,2p\}$.

Note that if $\operatorname{ord}_r(a)=2$, then $r|a^2-1 \equiv (11^2-17^2)b^2 \pmod r$, which gives $r=7$ as the only possibility. On the other hand, $\operatorname{ord}_r(a)=2p$ implies $2p|r-1$. Thus, all prime divisors of $3p^{q-1}+1$ other than $2$ or $7$ are congruent to $1$ modulo $2p$, i.e.
$$3p^{q-1}+1=2^{\alpha}7^{\beta}p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_k^{\gamma_k}, \quad (*)$$
where $p_i \notin \{2,7\}$ are prime divisors with $p_i \equiv 1 \pmod{2p}$.

$$\frac{11^p+17^p}{28}=11^{p-1}-11^{p-2}17+11^{p-3}17^2-\dots+17^{p-1} \equiv p4^{p-1} \pmod 7,$$
so $11^p+17^p$ is not divisible by $7^2$ and hence $\beta \le 1$.
If $q=2$, then $(*)$ becomes $3p+1=2^{\alpha}7^{\beta}p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_k^{\gamma_k}$, but $p_i \ge 2p+1$, which is only possible if $\gamma_i=0$ for all $i$, i.e. $3p+1=2^{\alpha}7^{\beta} \in \{2,4,14,28\}$, which gives us no solutions.
Thus $q>2$, which implies $4|3p^{q-1}+1$, i.e. $\alpha=2$. Now the right hand side of $(*)$ is congruent to $4$ or $28$ modulo $p$, which gives us $p=3$. Consequently $3^q+1 \equiv 6244 \pmod{3}$ which is only possible for $q=3$. The pair $(p,q)=(3,3)$ is indeed a solution.