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I am seeking sufficient conditions to ensure that a monic polynomial, denoted as $f$ in $\mathbf{Z}[x]$, possesses exclusively simple roots.

Based on an old paper (this reference), it has been established that polynomials of the form $f(z)=1+z^p+z^q-z^n$ where $1\leq p< q <n$ exhibit simple roots (See this post).

Q. What other general forms of such monic polynomials can be considered?

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    $\begingroup$ In general, this is governed by the (non)vanishing of the discriminant, which is a degree $2n-2$ expression in the coefficients of $f$ that can be computed by a procedure comparable to long division. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Nov 9, 2023 at 12:16
  • $\begingroup$ That sounds nice! Could you please give some more details about. $\endgroup$
    – ABB
    Commented Nov 9, 2023 at 12:53
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    $\begingroup$ The question is far to vague to expect a useful answer. For instance, every polynomial which is irreducible, for instance because it is Eisenstein, has no multiple roots. $\endgroup$
    – Peter Mueller
    Commented Nov 9, 2023 at 13:49

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Discriminant is the resultant of $f$ and $f'$. In your case, it is simply an integer. If this integer is $\neq 0$ then $f$ has no multiple root (and vice versa).

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