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If we substitute $y:=v+n$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{n-k}_q \, .$$

In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-z}\cdot f_{x+v}(qz)=f_{x+v+1}(z)\, ,$$ which is straightforward.

($\prod_{k=v+1}^{v+1+x}\,\prod_{k=0}^{v}=\prod_{k=1}^{v+1+x}\,\prod_{k=0}^{0}=\prod_{k=0}^{v+x+1}$)

If we substitute $y:=v+n$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{n-k}_q \, .$$

In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-z}\cdot f_{x+v}(qz)=f_{x+v+1}(z)\, ,$$ which is straightforward.

($\prod_{k=v+1}^{v+1+x}\,\prod_{k=0}^{v}=\prod_{k=1}^{v+1+x}\,\prod_{k=0}^{0}=\prod_{k=0}^{v+x+1}$) $$*$$ rmk. Note that the latter also gives again the first identity of Gjergji's answer, since $[z^n]f_{x+v+1}(z)=\binom{ x+v+1+n}{n}_q=\binom{ x+y+1}{n}_q$.

If we substitute $y:=v+nw$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{n-k}_q \, .$$

In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-z}\cdot f_{x+v}(qz)$$ which is straightforward.

If we substitute $y:=v+n$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{n-k}_q \, .$$

In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-z}\cdot f_{x+v}(qz)=f_{x+v+1}(z)\, ,$$ which is straightforward.

($\prod_{k=v+1}^{v+1+x}\,\prod_{k=0}^{v}=\prod_{k=1}^{v+1+x}\,\prod_{k=0}^{0}=\prod_{k=0}^{v+x+1}$)

If we substitute $y:=v+nw$ in the identity, the LHS becomes a convolution of two similar sequences. In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-qz}\cdot f_{x+v}(qz)$$ which is straightforward.

If we substitute $y:=v+nw$ in the identity, the LHS becomes a convolution of two similar sequences, $$\sum_{k=0}^nq^{(v+1)k}\binom{x+k}{k}_q \binom{v+(n-k)}{n-k}_q =\sum_{k=0}^n q^{n-k}\binom{x+v+n-k}{n-k}_q \, .$$

In terms of the known series $$f_x(z):=\sum_{k=0}^\infty \binom{x+k}{k}_q \,z^k=\prod_{k=0}^x{1\over1-q^kz}$$ the identity reads $$f_x(q^{v+1}z)\cdot f_v(z)={1\over1-z}\cdot f_{x+v}(qz)$$ which is straightforward.