Because of $\overline{\cosh(z)}=\cosh(\bar{z})$ it can be produced by

In[23]:= Simplify[TrigExpand[TrigReduce[Cosh[x + I y] Cosh[x - I y]]] /. {Sin[y]^2 -> (1 - Cos[y]^2),
                                                                          Cosh[x]^2 -> (1 + Sinh[x]^2)}]
Out[23]= Cos[y]^2 + Sinh[x]^2

and then checked again by

In[37]:= FullSimplify[Cosh[x + I y] Cosh[x - I y] - (Cos[y]^2 + Sinh[x]^2)]
Out[37]= 0

B.t.w. there are equivalent answers, e.g. if one takes the other side of the basic identies in replacement

In[39]:= Simplify[TrigExpand[TrigReduce[Cosh[x + I y] Cosh[x - I y]]] /. {Cos[y]^2 -> (1 - Sin[y]^2), 
                                                                          Sinh[x]^2 -> (Cosh[x]^2 - 1)}]
Out[39]= Cosh[x]^2 - Sin[y]^2

In[41]:= FullSimplify[Cosh[x + I y] Cosh[x - I y] - Cosh[x]^2 + Sin[y]^2]
Out[41]= 0

The equality looks false:

Plot[{Abs[Cosh[x]]^2, Sinh[x]^2 + Cos[x]^2}, {x, -1, 1}]