Polynomial decomposition test
James Rickards answers this question with his paper "When Is a Polynomial a Composition of Other Polynomials?" in The American Mathematical Monthly (118:4, pp. 358-363). In particular, a polynomial of degree $mn$ can be written as a composition of polynomials of degree $m$ and $n$ exactly when the roots of the polynomial can be partitioned into multisets $S_1,\ldots,S_m$ with $n$ elements each such that, for $0<j<n,$ $$ \sigma_j(S_1)=\cdots=\sigma_j(S_m) $$ where $\sigma_j$ is the $j$th symmetric polynomial.
If $f(x)=g(h(x))$ then $\deg f=\deg g\deg h$, so you should factor $\deg f$ to look for candidate degrees.
If $\deg f$ is prime, the anwer is trivially "no".
Another case where the decomposition is obvious is when all monomials in $f$ have exponents a multiple of some $d$; then $h(x)=x^d$ works.
It may be worthwhile to apply a linear substitution to $f$ to get rid of the second-highest power of $x$; maybe that produces case 2.
To check for quadratic $h$, a suitable linear substitution would make $h$ an even function; then $f$ becomes even as well (which is in fact contained in case 2 above; quadratic $h$ can always be assumed to be just $x^2$ after linear substitution)
To check for cubic $h$, a suitable linear substitution plus vertical offset (that can be eaten up by $g$) makes $h$ and hence also $f$ odd; the substitution is again the one from 2 and must produce only odd exponents. For small degrees, the conditions on the coefficients may be checked manually.
In all of the above, roots of $h$ are also roots of $f$, so finding some of the latter and trying to combine some $h$ from some of them may be promising.
A different approach is to note that $f'(x)=h'(x)g'(h(x))$ so that any roots of $h'$ are also roots of $f'$.