I claim the answer is 16. In what follows, I assume $\mathbb{N}=\{1,2,\dots\}$, namely not including zero. Note that, for $(m,n)=(4,2)$, $16$ is indeed achievable. I know prove this is the smallest possible. We first investigate $f(m,n)\triangleq 3\cdot 5^m-11\cdot 13^n$. Note that since $f$ is even, and $f(m,n)\equiv 1\pmod{3}$, $f(m,n)\in\{4,10\}$ are the only cases to rule out. Observe that $f(m,n)\equiv 0\pmod{4}$, hence $f(m,n)=10$ is impossible. We now study $f(m,n)=4$. In this case, modulo $13$ yields that $5^m\equiv -3\pmod{13}$. However, $5^m\in\{5,-1,-5,1\}\pmod{13}$. This contradiction shows $f(m,n)=4$ is impossible, thus $f(m,n)\geqslant 16$, as I claimed. We now study $g(m,n)\triangleq -f(m,n)=11\cdot 13^n-3\cdot 5^m$. Analogous to above, $g(m,n)\equiv 2\pmod{3}$, and $4\mid g(m,n)$. Thus, the only possible case to rule out is $g(m,n)=8$. Notice now that modulo $13$ again yields $-3\cdot 5^m\equiv 8\pmod{13}\iff 5^m\equiv 6\pmod{13}$. As I have proven above, however, this is impossible. Thus, $16$ is indeed the smallest possibility.