Why is int only 2 bytes?
The ATmega328 used in a lot of Arduinos is a 8-bit microcontroller. This means that registers are 8-bits, the data bus is 8-bits, the ports are 8-bits. There are some minimal 16-bit aspects to the system (e.g. one of the timers), but nearly everything is 8-bits.

Therefore, most operations handle 8-bits at a time. Working on anything except 8-bits (i.e. 16-bit or 32-bit integers and floating point numbers) requires what could essentially be described as software emulation, where the compiler uses multiple instructions to work on these larger variables.

8-bits is obviously adequate to address a 8-bit port. It's also enough to deal with many loop counters, return values, and ASCII characters. It isn't really enough though when dealing with numbers. A signed 8-bit int (int8_t) can only represent -128 -> +127. Unsigned (uint8_t) can only represent 0 -> 255.

8-bit integers are quite limiting. C/C++ int must represent at least -32,678 -> +32,767 so maps to int16_t - the smallest size that will do so. This gives a good balance of range and efficiency. This is especially important when beginners are learning - overflow is not really something that non-programmers understand.

There is a performance impact of doing this however, because most 16-bit operations take at least twice as long as an 8-bit operation, and use twice as many registers. This may or may not make a difference to you.

Many of us switch to the native types such as int8_t and uint8_t as it gives you far more control.