What will happen if I use RSA encryption for plaintext larger than RSA bit size? I have often read that using RSA for encrypting bigger size plaintext is not good solution.

What I understand is that using RSA encrytion is best for small plaintext and most effective if it is smaller than 'bit size' of RSA. For instance, if I use RSA with a 2048 bit size then the plaintext that I can encrypt must be smaller than 256 bytes.

Is this true? What happens if the plaintext is bigger than 256 bytes? Why is there a problem with larger plaintext when performing RSA encryption?
If you have an RSA public key $(e, N)$ (in the case of RSA2048 $N$ will be 2048 bits) and private key $d$ then you can only encrypt messages $m$ whose integer representation is less than $N$.

Recall that in RSA encryption is defined as $c = m^e \text{ mod } N$ where $c$ is the resulting ciphertext, and decryption is defined as $m = c^d \text{ mod } N$. The key insight is that decryption cannot produce an $m \ge N$ because of the modulo $N$ operation (the remainder of division by $N$ must be less than $N$). Thus if we encrypt an $m \ge N$ decryption will not produce the correct value.

To get around this it is possible to break $m$ up into chunks that are smaller than $N$, encrypt the chunks, and then recombine the chunks after decrypting to form $m$. For example ($||$ denotes concatenation):

$$m = m_1 || m_2$$ $$c1, c2 = Enc(m_1), Enc(m2)$$ $$m = Dec(c_1) || Dec(c_2)$$

More things to consider beyond this are space that padding will take (unpadded RSA is not secure), and whether you should be using a symmetric cipher like AES which is more suited for encrypting large amounts of data (RSA is much slower than symmetric ciphers and was not designed to encrypt large amounts of data).

Edit: If we do encrypt a message $m \ge N$ the ciphertext $c$ resulting from encrypting $m$ will not decrypt back to $m$. Example using some small parameters:

$$p =17, q = 19, N = p*q = 323, e = 5, d = 173$$
$$m = 400$$
$$Enc(m) = c = 400^5 \text{ mod } 323 = 229$$
$$Dec(c) = m' = 229^{173} \text{ mod } 323 = 77$$

Note that $m \ne m'$ and thus for values $m \ge N$ RSA does not decrypt properly. In fact, for any value $m \ge N$ encrypted via RSA the ensuing decryption will produce $m \text{ mod }N$ instead of $m$ (as is the case in the example: $77 = 400 \text{ mod } 323$).