Find $a, b, c \in \mathbb{N}$ such that $11^{a}+3^{b}=c^{2}$.
By looking modulo 3, we find that $a$ is even. Then, by setting $a=2 a^{\prime}$ and with some algebraic manipulations,

$$
3^{b}=\left(c-11^{a^{\prime}}\right)\left(c+11^{a^{\prime}}\right)
$$

Thus, by using the prime factorization, $c-11^{a^{\prime}} \mid c+11^{a^{\prime}}$ which means $c-11^{a^{\prime}} \mid 2 \cdot 11^{a^{\prime}}$.

But since $c-11^{a^{\prime}}=3^{\beta}$, we get $\beta=0$, so $c-11^{a^{\prime}}=1$ and $c+11^{a^{\prime}}=3^{b}$.

We subtract and find then $3^{b}-1=2 \cdot 11^{a^{\prime}}$. We look at the order of 3 modulo 11 and find that it divides $b$ and is 5, so $3^{5 b^{\prime}}-1=2 \cdot 11^{a^{\prime}}$. Here we can use Zsigmondy, because $3^{5}-1=2 \cdot 11^{2}$. If we take a $b^{\prime} \geqslant 2$, we will have a prime factor in the left term that is neither 2 nor 11, so there are no other solutions than $(4,5,122)$.