If $R$ is a UFD, then every prime element of $R$ generates a prime ideal

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The hint in my book says:

If $p$ is prime element of $R$ $\implies$ $p$ is irreducible element of $R$ $\implies$ $\langle p\rangle$ is maximal ideal of $R$ $\implies$ $\langle p\rangle$ is prime ideal of $R$.

I am stuck with the second implication

$p$ is irreducible element of $R$ $\implies$ $\langle p\rangle$ is maximal ideal of $R$

Being a beginner, I know this holds for a $PID$, but could not proceed for a $UFD$. Some reasoning for the above implication would be very helpful.

NOTE: $\langle p\rangle$ denotes the principal ideal generated by $p$.