Consider the group $G(n)= \langle a, b \ \vert\ aba^{-1}=b^{n+1}, bab^{-1}=a^{n+1} \rangle$, $n\ge 1.$
Show that the center $Z$ is cyclic of order $n$ and that $G/Z$ is abelian of order $n^2$.
This group is somewhat related to the group mentioned in 479835.
It is easy to see that the commutator quotient $G/G'$ is isomorphic to a direct product of two cyclic groups of order $n$. So the problem is equivalent to showing $Z=G'$.
Note that $b^{n} = a b a^{-1} b^{-1} = (b a b^{-1} a^{-1})^{-1} = a^{-n}$. So $b$ and $a$ commute with the commutator $[a, b] = a b a^{-1} b^{-1}$, which is therefore central, and so $a^{n}$ and $b^{n}$ are also central. It follows that $$ 1 = [a^{n}, b] = [a, b]^{n}, $$ so $[a,b]$ has order at most $n$. Clearly $G' = \langle [a, b] \rangle$.
Addendum
Now to show that $G'$ has order $n$, one need to construct an example, as suggested by Derek Holt. Let us do it for $n$ odd. (Thanks i. m. soloveichik for the comment.)
I guess it goes like this. Start with a group $\langle a \rangle$ of order $n^{2}$, and consider its automorphism $c$ of order $n$ which sends $a$ to $a^{1+n}$. Consider the semidirect product $\langle a \rangle \rtimes \langle c \rangle$, and in it $b = a^{-1} c$. These elements should satisfy the given relations.
Completing the calculations
In fact by definition we have $c a c^{-1} = a^{1+n}$. Then $$ b a b^{-1} = a^{-1} c a c^{-1} a = a^{1+n}, $$ and $$ a b a^{-1} = a a^{-1} c a^{-1} = c a^{-1} c^{-1} a = a^{-1-n} c. $$ Now $$ b^{n} = (a^{-1} c)^{n} = [c, a^{-1}]^{\binom{n}{2}}a^{-n} c^{n} = a^{-n}, $$ as $n$ divides $\dbinom{n}{2}$, and $[a^{-1}, c] = a^{n}$, of order $n$. This $b^{n + 1} = a^{-1-n} c$.