Write the general term of $\left(x^m+\frac1{x^2}\right)^{22}.$

The general term is $T_{r+1}=\binom{22}{r}(x^m)^{22-r}\left(x^{-2}\right)^r,$ where $r=0,1,2,\dots,22$.

Simplify the power of $x$:

We need the coefficient of $x$, i.e. the coefficient of $x^1$. So the exponent must satisfy $22m-r(m+2)=1.$

Rearranging, $r(m+2)=22m-1.$

The coefficient of this term is then $\binom{22}{r}=1540.$

Now find $r$ such that $\binom{22}{r}=1540.$

We know $\binom{22}{3}=\frac{22\cdot21\cdot20}{3\cdot2\cdot1}=1540.$ Also, by symmetry, $\binom{22}{19}=1540.$

So possible values are $r=3 \quad \text{or} \quad r=19.$

Use the exponent condition

Case 1: $r=3$

$22m-3(m+2)=1$ $22m-3m-6=1$ $19m=7$ $m=\frac{7}{19},$ which is not a natural number.

Case 2: $r=19$

$22m-19(m+2)=1$ $22m-19m-38=1$ $3m=39$ $m=13.$

This is a natural number.

Therefore, the required natural number is $\boxed{13}.$