Maximum Number of Blocks to be Encrypted under One Key in CBC and CTR Mode?
CTR mode.
There's a generic birthday attack on CTR mode of a block cipher.

The idea of CTR mode is to use a PRF, or pseudorandom function family, $F_k\colon \{0,1\}^\iota \to \{0,1\}^n$ to approximate a one-time pad with a short key $k$: $$c = m \oplus (F_k(0) \mathbin\| F_k(1) \mathbin\| F_k(2) \mathbin\| \cdots \mathbin\| F_k(\ell - 1)),$$ where $\ell$ is the length in $n$-bit blocks of the message $m$ (ignoring details of partial blocks).

If instead of a PRF we have a PRP, a pseudorandom permutation family or $n$-bit block cipher, $E_k\colon \{0,1\}^n \to \{0,1\}^n$, we can use that to approximate a PRF.  But a PRP, being a family of permutations, has the property that in the sequence of blocks $E_k(0), E_k(1), \dots, E_k(\ell - 1)$, as long as $\ell &lt; 2^n$, there are no repeated blocks.  In a uniform random sequence of blocks, we would expect to see a collision when $\ell \approx \sqrt{2^n} = 2^{n/2}$, by the birthday paradox.

When $n = 128$ as in AES, this procedure—check for a repeated block—enables an attacker to distinguish a very long but not unimaginably long CTR mode pad from a very long uniform random pad with nonnegligible probability.

CBC mode.
In CBC mode, the $i^\mathit{th}$ ciphertext block is $c_i = E_k(m_i \oplus c_{i-1})$, with $c_{-1}$ being the initialization vector.  If the adversary notices that $c_i = c_j$ then they can infer $$m_i \oplus c_{i-1} = m_j \oplus c_{j-1}$$ from which they learn $$m_i \oplus m_j = c_{i-1} \oplus c_{j-1}.$$  Turning this into a distinguisher and calculating how the birthday bound is relevant here is left as an exercise for the reader who has had more breakfast so far this morning than I have.

Recommendations.
Because of these birthday attacks, cryptographers usually sternly advise against using AES-CTR or AES-CBC for more than a few exabytes of data, or, to be conservative, a few hundred terabytes.  (It doesn't matter whether that's for a lot of short messages or a few long messages.)  64-bit block ciphers like 3DES fare even worse, to the point that they can be broken in practice.

When $n = 256$ as in Threefish-256, these attacks are irrelevant: the birthday bound is around $2^{128}$ blocks of data, which you will never see.  When it's a PRF with $n = 512$, as in Salsa20 or ChaCha, it's even less relevant—although if you pick per-message nonces at random, the birthday bound once again becomes relevant because those nonces are only 64 bits and there is a nonnegligible chance of a security-destroying collision in the nonce, which is why the extended-nonce variants XSalsa20 and XChaCha exist and why some cryptographers and standardizers are considering nonce-misuse-resistant AEAD schemes.   In sum, the birthday paradox is a versatile tool!