There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical work of Thue (1909).

Assume that $2^j3^k+1=n^2$ is a square. Then $n-1=2^a3^b$ and $n+1=2^c3^d$ for some $a,b,c,d\in\mathbb{N}$. Classifying the quadruple $(a,b,c,d)$ according to its residue modulo $3$, we are led to $81$ equations of the form $Ax^3-By^3=2$, where the coefficients $A$ and $B$ lie in $\{2^p3^q:p,q\in\{0,1,2\}\}$. Each of them has finitely many solutions by a theorem of Thue (1909), so we are done. See also Thue equations.

There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical work of Thue (1909).

Assume that $2^j3^k+1=n^2$ is a square. Then $n-1=2^a3^b$ and $n+1=2^c3^d$ for some $a,b,c,d\in\mathbb{N}$. Classifying the quadruple $(a,b,c,d)$ according to its residue modulo $3$, we are led to $81$ equations of the form $Ax^3-By^3=2$, where the coefficients $A$ and $B$ lie in $\{2^p3^q:p,q\in\{0,1,2\}\}$. Each of them has finitely many solutions by a theorem of Thue (1909), so we are done. See also Thue equations.

P.S. This method generalizes to arbitrary many primes instead of just $2$ and $3$. For the original problem at hand, an elementary treatment fits better. See Fedor Petrov's answer.

There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical work of Thue (1909).

Assume that $2^j3^k+1=n^2$ is a square. Then $n-1=2^a3^b$ and $n+1=2^c3^d$ for some $a,b,c,d\in\mathbb{N}$. Classifying the quadruple $(a,b,c,d)$ according to its residue modulo $3$, we are led to $81$ equations of the form $Ax^3-By^3=2$, where the coefficients $A$ and $B$ lie in $\{2^p3^q:p,q\in\{0,1,2\}\}$. Each of them has finitely many solutions by a theorem of Thue (1909), so we are done.

There are finitely many pairs $(j,k)$, and this follows from results on $S$-unit equations. Moreover, the solutions can be effectively determined. Here is a quick treatment going back to the classical work of Thue (1909).

Assume that $2^j3^k+1=n^2$ is a square. Then $n-1=2^a3^b$ and $n+1=2^c3^d$ for some $a,b,c,d\in\mathbb{N}$. Classifying the quadruple $(a,b,c,d)$ according to its residue modulo $3$, we are led to $81$ equations of the form $Ax^3-By^3=2$, where the coefficients $A$ and $B$ lie in $\{2^p3^q:p,q\in\{0,1,2\}\}$. Each of them has finitely many solutions by a theorem of Thue (1909), so we are done. See also Thue equations.