Did you just simply go through several test cases with a bit of thinking to get that? Or was there some method, which generates several more solutions, or which gave that one (which may be the only one)?

We can prove the next result: For any positive integers $\left( {\alpha ,\beta } \right) = 1$, there exist positive integers $m,n$, such that ${m^\alpha } + {n^\beta }$ is a squre. Proof: Suppose $\beta $ is odd, then $(2\alpha ,\beta ) = 1$, so there exist positive integers $r,s$，such that $\beta s - 2\alpha r = 1$. For $m = {3^{2r}}$, $n = {3^s}$, ${m^\alpha } + {n^\beta } = {3^{2\alpha r}}(1 + 3) = {\left( {2 \times {3^{\alpha r}}} \right)^2}$ is a square.

And further on, one can take any number of the form $k^2 -1$ instead of $3$.

See also here. Also, I was forwarded a scanned copy of the test, on which the problem appears to have asked instead whether $m^{20} + 11^n$ could ever be a perfect square, with $m,n>0$.

Assume $m^{20} + 11^n = a^2$, so $11^n = (a-m^{10})(a+m^{10})$. Then $a-m^{10} = 11^{\alpha}$ and $a+m^{10} = 11^{\beta}$, with $0 \leq \alpha < \beta$ and $\alpha+\beta = n$. But then $2m^{10} = 11^{\alpha}(11^{\beta - \alpha} - 1)$. If $\alpha = 0$, this leads to $-1 \equiv 2m^{10} \equiv 2 \pmod{11}$ (since $\gcd(m,11)=1$), absurd. If $\alpha > 0$ we need $11 \mid m$, and so $2\cdot 11^{\gamma}\ell^{10} = 11^{\beta - \alpha} - 1$ for some $\gamma \geq 0$ and $\gcd(\ell,11) = 1$. This implies $\gamma = 0$, hence $2\ell^{10} = 11^{\beta - \alpha} - 1$, without solution, since it reduces to the case before. Therefore no solutions.