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Reduce coefficients modulo $I$:This is a surjective ring homomorphism and its kernel consists exactly of polynomials all of whose coefficients lie in $I$, namely $I[X]$. The first isomorphism theorem gives the coefficientwise quotient of a polynomial ringIf $I$ is prime, then $R/I$ is an integral domain, so $(R/I)[X]$ is an integral domain. The quotient criterion therefore shows that $I[X]$ is prime in $R[X]$.

Pass to $A=R/I$. It is enough to prove that the zero ideal of $A[t]$ is primary. The zero ideal of $A$ is primary, so every zero divisor of $A$ is nilpotent element.

Suppose $fg=0$ in $A[t]$ with $g\ne 0$. By McCoy theorem, some nonzero $a\in A$ satisfies $af=0$. Hence every coefficient of $f$ is a zero divisor and therefore nilpotent. There are only finitely many coefficients, so the ideal they generate is nilpotent; consequently some power of $f$ is zero. This proves that $(0)$ is primary in $A[t]$, and the coefficientwise quotient of a polynomial ringshows that $I^e$ is primary.