We have $n>1$. If $ m\in \{1,2,3,4,5,6 \} $ then solution is only $m=3$, $n=2$. So we solve at $m\ge 7$ and $n\ge 2$. (1). Suppose that $n=2^b(b\ge 1) $. Then since $ 7|m!=2^n+n=2^x+2^y\equiv \{ 1,2,4 \} +\{ 1,2,4 \} $ we get contradiction. (2). Suppose that $n=2^b\cdot k, (b\ge 1,(k>1) - odd)$. If $k\le m$ then $k|m!$ and $k|2^n$ - contradiction. If $k\ge m$ then since $ a=v_2(m!) $ and $ \lfloor \frac{m}{2} \rfloor \le a \le m $ and $m\ge 7$ we get $2^n >2^{m^2}>m^m>m!$ and contradiction. Answer: $(m,n)=(3,2)$.

I didn't understood the part $ k\ge m $, why $n>m^2$ ?

why $n>m^2$?

sorry, ok @rightways; We have $v(2^n+n)=b\ge \lfloor \frac{m}{2} \rfloor $ Those not $m^2$ $ \rightarrow m\cdot \lfloor \frac{m}{2} \rfloor $. But, it's correct, because $m\ge 7.$

of course, we have $2^{\lfloor \frac{m}{2}\rfloor}\ge m$ for any $m\ge 7$.

Assume that $m\ge4$ and $n\ge2$.First note that n must be even.If n is of form 4k+2 then $2^n+n \equiv 2\bmod{4}$. But $4|m!$ and hence there is no solution. Hence $4|n$ Case-1 $n \equiv 0\bmod{12}$ In this case, $2^n+n=2^{12k}+12k\equiv 1\bmod{3}$ But $3|m!$.No solution. Case-2 $n\equiv 4\bmod{12}$ In this case, $2^n+n=2^{12k+4}+12k+4\equiv 2\bmod{3}$ No solution. There's only one case left, when $n\equiv 8\mod{12}$ I will be grateful to the person who solves this case.Please try.

Sayantanchakraborty, sorry to say this but your solution is nothing. Because we obviously know $n=p-1(modp)$ will satisfy $(modp)$ so you cant get something from mods. (but i noticed that after exam:()

Find all positive integer (x, y) such that  2(x^3-x) =y^3-y

Try the base case when m<=6 the solution is (m,n)=(3,2) We have a lemma: v2(m!)=m-s2(m) (s2 is the sum of digit of n written in binary). Note that 2^m+m<m! for m>=4 which can prove easily by induction m. From the condition of the problem we have n>m and from v2(2^n)=n so it equivalent to v2(m!-n)=n but v2(m!)=m-s2(m)<m-1 by the lemma Written n=2^(m-s2(m)).k (k is odd) then if k<m then m!-n divisible by k and 2^n divisible by k (contradiction). So k>=m hence n<m which is false because n>m

If $m \geq 2$ $\implies n$ is even number.$n=2k$. $$2^{2k} + 2k \geq (2^k + 1)^2 =2^{2k}+ 2^{k+1} + 1$$$$2k \geq 2^{k+1}+1$$Remaining easy.$\blacksquare$

Answer: $(m,n)=(3,2)$. $\textbf{Claim:} \ 2^{2^x}>(x+1)!$ Proof: It's true for $x=1,2$. If inequality holds for $x=k,$ then $2^{2^{k+1}}=(2^{2^k})^2>(k+1)!^2>(k+2)!$ since $(k+1)!\geq k(k+1)>(k+2)$ for $k\geq 2$. $\square$ Denote $n=2^\alpha k$ where $k$ is odd. Let $m=2^c+b$ where $b\in [0,2^c)$. Suppose that $k\geq 3$. If $p|k$ and $p<m,$ then $p|m!=2^\alpha(2^{2^\alpha k-\alpha}+k)\implies p|2$ which is impossible. Thus, $p|k\implies p>m$ so $k>m$. $\alpha=v_2(m!)\geq m-\lceil log_2(m) \rceil\geq 2^c+b-c-1$ \[(2^c+b)!=2^{2^\alpha k}+2^\alpha k>2^{2^\alpha m}\geq 2^{2^{2^c+b-c-1} m}\geq 2^{2^{2^c+b-c-1}.2^c}=2^{2^{2^{c}+b-1}}>(2^c+b)!\]Which gives a contradiction. Thus, $k=1$. \[2^{\alpha}(2^{2^\alpha-\alpha}+1)=m!\]If $m\geq 5,$ then $5|2^{2^\alpha-\alpha}+1\implies 2^\alpha-\alpha\equiv 2(mod \ 4)\implies \alpha$ is even. $3|2^{2^\alpha-\alpha}+1\implies \alpha$ is odd. These two contradict hence $m\leq 4$. There is no solution for $m=1,2,4$ and $m=3$ gives $n=2$ as desired.$\blacksquare$