I got $(2,3)$ as a solution.

Clearly $n=2$ gives $k=3$, whereas $n=1$ gives nothing. Assume $n\ge 3$. Check that $n^3-5n+10\in\{3,6,1,1,5,5,0\}\pmod{7}$ whereas $2^k\in\{2,4,1\}$. Thus, $3\mid k$. Now that we find right hand side to be a cube, so is $n^3-5n+10$. Check that $n^3-5n+10<n^3$ for $n>2$. Moreover, it is not hard to see $(n-1)^3 =n^3-3n^2+3n-1<n^3-5n+10$. Thus we squeezed this object between two consecutive cubes. Hence no solutions except $n=2$.

how did you came up with mod 7 for bounding.

Cubes play nicely in modulo 7.

If $n=1$, $1-5+10 = 6 \ne 2^k$ for any positive integer $k$. If $n=2$, $8-10+10 = 8 = 2^3$, so $(n,k) = (2,3)$ is a solution. Let $f(n) = n^3 - 5n + 10$. Since $f(n) = 2^k$, $f(n)$ must be even. When $n$ is even, say $n=2m$, then \[ (2m)^3 - 5(2m) + 10 = 8m^3 - 10m + 10 \equiv -2m+2 \equiv 0 \pmod{4}. \]Thus $2m \equiv 2 \pmod{4}$, which implies $m \equiv 1 \pmod{2}$. Therefore, $m$ is odd, say $m=2a+1$ for some integer $a \ge 0$. Then $n = 2m = 2(2a+1) = 4a+2$. When $n$ is odd, say $n=2m+1$, then \begin{align*} (2m+1)^3 - 5(2m+1) + 10 &= 8m^3 + 12m^2 + 6m + 1 - 10m - 5 + 10 \\ &= 8m^3 + 12m^2 - 4m + 6 \equiv 2 \pmod{4} \end{align*}This is not possible since $f(n) = 2^k$ is divisible by 4 if $k \ge 2$. Thus, $n$ must be even. We already know that $n$ must be of the form $4a+2$. We found $n=2$, $k=3$. Suppose $n > 2$. When $n=2$, $2^3 - 5(2) + 10 = 8 = 2^3$. If $k=1$, $n^3-5n+10=2$, $n^3-5n+8=0$. No integer solutions. If $k=2$, $n^3-5n+10=4$, $n^3-5n+6=0$, $(n-2)(n^2+2n-3)=0$, $(n-2)(n-1)(n+3)=0$. $n=1,2,-3$. Only $n=2$ is positive. $2^3-5(2)+10 = 8=2^3$, $k=3$. If $k=3$, $n^3-5n+10=8$, $n^3-5n+2=0$. We know $n=2$ is a root. $(n-2)(n^2+2n-1)=0$. $n=2$, $n=-1\pm\sqrt{2}$. So $n=2$. When $n=2$, $2^3-5(2)+10 = 8 = 2^3$. So $(n,k)=(2,3)$ is a solution. Let $k > 3$. Then $n$ must be even. Let $n=2m$, $8m^3 - 10m + 10 = 2^k$. $4m^3 - 5m + 5 = 2^{k-1}$. Since $k > 3$, $k-1 > 2$. So $4m^3 - 5m + 5$ is divisible by 4. $4m^3 - 5m + 5 \equiv -m+1 \equiv 0 \pmod{4}$. $m \equiv 1 \pmod{4}$, $m=4a+1$ for some $a \ge 0$. Then $n=2m=8a+2$. $f(n=2) = 8 = 2^3$. Suppose $n>2$, then $k > 3$. Therefore, the only solution is $\boxed{(2,3)}$.

Cubes play nicely in modulo 7.

Cubes play nicely in modulo 7.