Note that $\mod 3$ gives $m$ is even and $\mod 4$ gives $n$ is even. We get $$2^{2k}= (X^2-3^l)(X^2+3^l)\implies X^2-3^l= 2^a,~~ X^2+3^l=2^b \implies 2^a| 2^b \implies 2^a|2\cdot 3^l| \implies a=0 \text{or} 1$$So $X^2-3^l=1, 2$ but squares are $0,1\mod 3.$ So $$X^2-3^l=1 \mod 3$$Now we will try to solve $$X^2-3^l=1$$ Note that $$X^2-1=(X-1)(X+1)=3^l\implies X-1=3^w,X+1=3^v\implies 3^w|2\implies w=0 \implies X-1=1\implies 2.$$ Clearly, $X=2$ doesn't satisfy. So we don't have any solutions.