let x = [];

What is the content of x? If you said an empty array you would be wrong, strictly speaking. The right answer is that x contains a reference (e.g. <#525>) which points to a memory location which contains the value [].

let x = 5;

What is the content of x here? You read the previous paragraph and say: It contains a reference which points to a memory location which contains the value 5? That would be wrong. This time it does simply contain the value 5, no references included.

Why? What is so different between [] and 5, is that [] is an object data type and 5 is a primitive data type.

JavaScript has five primitive data types: Boolean, null, undefined, String, and Number. The rest are objects, including Array, Function and Object.

Primitive types are copied by value.

let x = 1; // x contains 1
let y = x; // y contains copy (not reference) of x which is 1
y = 2; // overwriting with 2 does not affect x in any way

console.log(x === 1); // true
console.log(y === 2); // true

Objects are copied by having their reference copied:

const obj = {
    innerObj: {
        x: 9
    }
};

const z = obj.innerObj;
z.x = 25;
console.log(obj.innerObj.x); // 25

With const z = obj.innerObj; z contains a reference which points to the memory location which stores the value of innerObj.

const obj = {
    arr: [{ x: 17 }]
};
 
let z = obj.arr;
 
z = [{ x: 25 }];
 
console.log(obj.arr[0].x); // 17

With z = [{ x: 25 }]; we overwrote the reference that z contained before. It is now unreachable and gets garbage collected by JavaScript.

const obj = {
    arr: []
};
 
obj.arr.push(17);
 
console.log(obj.arr === [17]);  // false

obj.arr does contain [17] but not the [17] that we are comparing with in the last line.